IN  MEMORIAM 
FLOR1AN  CAJORI 


PLANE    GEOMETRY 


BY 


EDWARD    RUTLEDGE   ROBBINS,    A.B. 

SENIOR    MATHEMATICAL    MASTER 
THE    WILLIAM    PENN    CHARTER    SCHOOL 


NEW  YORK  .-CINCINNATI --CHIC AGO 

AMERICAN    BOOK    COMPANY 


COPYRIGHT,  1906,  BY 
EDWARD   RUTLEDGE  ROBBINS. 


PLANE    GEOMETRY. 
W.   P.    I 


FOR    THOSE    WHOSE    PRIVILEGE 
IT    MAY    BE    TO    ACQUIRE    A    KNOWLEDGE    OF 

GEOMETRY 

THIS   VOLUME    HAS   BEEN   WRITTEN 

AND    TO    THE    BOYS    AND    GIRLS    WHO    LEARN    THE    ANCIENT    SCIENCE 

FROM    THESE    PAGES,    AND    WHO    ESTEEM    THE    POWER 

OF    CORRECT   REASONING   THE    MORE 

BECAUSE    OF    THE    LOGIC    OF 

PURE   GEOMETRY 

THIS    VOLUME     IS     DEDICATED 


PREFACE 

THE  motives  actuating  the  author  in  the  preparation  of 
this  text  in  Geometry  have  been  : 

(a)    To  present  a  book  that  has  been  written  for  the  pupil. 

The  object  sought  in  the  study  of  Geometry  is  not  solely 
to  train  the  mind  to  accept  only  those  statements  as  truth 
for  which  convincing  reasons  can  be  provided,  but  to  culti- 
vate a  foresight  that  will  appreciate  both  the  purpose  in  mak- 
ing a  statement  and  the  process  of  reasoning  by  which  the 
ultimate  truth  is  established.  Thus,  the  study  of  this  formal 
science  should  develop  in  the  pupil  the  ability  to  pursue 
argument  coherently,  and  to  establish  one  truth  by  the  aid 
of  other  known  truths,  in  logical  order. 

The  more  mature  members  of  a  class  do  not  require  that 
the  reason  for  every  declaration  be  given  in  full  in  the  text ; 
still,  to  omit  it  altogether,  wrongs  those  pupils  who  do 
not  know  and  cannot  perceive  the  correct  reason.  But  to 
ask  for  the  reason  and  to  print  the  paragraph  reference  meets 
the  requirements  of  the  various  degrees  of  intellectual  capac- 
ity and  maturity  in  every  class.  The  pupil  who  knows  and 
knows  that  he  knows  need  not  consult  the  paragraph  cited  ; 
the  pupil  who  does  not  know  may  learn  for  himself  the 
correct  reason  by  the  reference.  It  is  obvious  that  the 
greater  progress  an  individual  makes  in  assimilating  the  sub- 
ject and  in  entering  into  its  spirit,  the  less  need  there  will 
be  for  the  printed  reference. 


6  PREFACE 

(6)    To  stimulate  the  mental  activity  of  tie  pupil. 

To  compel  a  young  student  to  supply  his  own  demonstra- 
tions, in  other  words,  to  think  and  reason  for  himself,  fre- 
quently proves  unprofitable  as  well  as  unpleasant,  and 
engenders  in  the  learner  a  distaste  for  a  study  he  has  the 
right  to  admire  and  to  delight  in.  The  short-sighted  youth 
absorbs  his  Geometry  by  memorizing,  only  to  find  that  his 
memory  has  been  an  enemy,  and  while  he  himself  is  becom- 
ing more  and  more  confused,  his  thoughtful  companion  is 
making  greater  and  greater  progress.  The  earlier  he  dis- 
covers his  error  the  better,  and  the  plan  of  this  text  gives 
him  an  opportunity  to  reestablish  himself  with  his  class. 
It  is  not  calculated  to  produce  accomplished  geometricians 
at  the  completion  of  the  first  book,  but  to  aid  the  learner 
in  his  progress  throughout  the  volume,  wherever  experience 
has  shown  that  he  is  likely  to  require  assistance.  It  is  cal- 
culated, under  good  instruction,  to  develop  a  clear  concep- 
tion of  the  geometric  idea,  and  to  produce  at  the  end  of  the 
course  a  rational  individual  and  a  friend  of  this  particular 
science. 

(c)  To  bring  the  pupil  to  the  theorems  and  their  demonstra- 
tions —  the  real  subject-matter  of  Geometry  —  as  early  in  the 
study  as  possible. 

(c?)  To  explain  rather  than  formally  demonstrate  the  simple 
fundamental  truths. 

(e)  To  apply  each  theorem  in  the  demonstration  of  other 
theorems  as  promptly  as  possible. 

(/)  To  present  a  text  that  will  be  clear,  consistent,  teach- 
able, and  sound. 


PRP:FACE  7 

% 

The  experienced  teacher  will  observe : 
(a)    The  economy  of  arrangement. 

Many  of  the  smaller  figures  are  placed  at  the  side  of  the 
page  rather  than  at  the  center.  The  individual  numbers  of 
theorems  are  omitted. 

(6)    The  superior  character  of  the  diagrams. 

(V)    The  omission  of  the  words  "  since  "  and  "for." 

The  advance  statement  is  made  and  the  reason  asked  for 
and  usually  cited.  The  inquiring  mind  fails  to  understand 
the  force  of  preceding  and  following  some  statements  with 
the  same  reason. 

(d)  Originals   that   are    carefully    classified*   graded*    and 
placed  after  the  natural  subdivisions  of  the  subject-matter. 

(e)  The  independence  of  these  originals. 

Every  exercise  can  be  solved  or  demonstrated  without 
the  use  of  any  other  exercise.  Only  the  truths  in  the 
numbered  paragraphs  are  necessary  in  working  originals. 

(/ )  The  setting  of  every  theorem*  corollary,  and  problem  of 
the  text  proper  infullface  type. 

(<7)  The  consistent  use  of  such  terms  as  " vertical  angles" 
" vertex-angle"  " adjacent  angles"  " angles  adjoining  a  side" 
and  others. 

(h)  The  full  treatment  of  measurement  and  the  illustrations 
of  the  terms  employed. 

(i)  The  summaries  that  precede  earlier  collections  of  origi- 
nal exercises. 

(/)  The  emphasis  given  to  the  discussion  of  original  oon- 
structions. 


8  PREFACE 

As  in  all  subjects  that  are  new  to  a  class,  the  successful 
teacher  will  be  content  with  short  lessons  at  the  beginning, 
and  will  progress  slowly  until  the  class  is  thoroughly  familiar 
with  the  language  and  the  general  method  and  purpose  of 
the  new  science. 

The  author  sincerely  desires  to  extend  his  thanks  to  those 
friends  who,  by  suggestion  and  encouragement,  have  inspired 
him  in  the  preparation  of  these  pages. 

EDWARD   R.   ROBBINS. 

THE  WILLIAM  PENN  CHARTER  SCHOOL, 
PHILADELPHIA. 


CONTENTS 

PAGE 

INTRODUCTION    .        .        .        . "      .        .        .        .        .        .        .11 

Angles 12 

Triangles 14 

Symbols        ...........  16 

Axioms .         .         .16 

Postulates 17 

BOOK  I.     ANGLES,  LINES,  RECTILINEAR  FIGURES          .         .        .  18 

Preliminary  Theorems          ........  18 

Theorems  and  Demonstrations    ...                  ...  20 

Model  Demonstrations 24 

Quadrilaterals 45 

Polygons       ...........  54 

Symmetry 58 

Locus 60 

Summary.     General  Directions  for  attacking  Originals     .         .  62 

Original  Exercises 64 

BOOK  II.     THE  CIRCLE 70 

Preliminary  Theorems 81 

Theorems  and  Demonstrations 82 

Summary      ...........  92 

Original  Exercises 93 

Kinds  of  Quantities.     Measurement  ......  97 

Original  Exercises 107 

Constructions       ..........  115 

Analysis 1-7 

Original  Constructions         ....                  ...  128 

9 


10  CONTENTS 

PAGE 

BOOK  III.     PROPORTION.     SIMILAR  FIGURES 140 

Theorems  and  Demonstrations   .......  141 

Original  Exercises  (Numerical)           ......  168 

Summary.     Original  Exercises  (Theorems)       ....  172 

Constructions       ..........  180 

Original  Constructions 184 

BOOK  IV.     AREAS 187 

Theorems  and  Demonstrations 187 

Original  Exercises  (Theorems) 197 

Formulas 201 

Original  Exercises  (Numerical)           ......  204 

Constructions .         .         .         .  207 

Original  Constructions         ........  213 

BOOK  V.    REGULAR  POLYGONS.    CIRCLES 217 

Theorems  and  Demonstrations    .         .         .         .        .         .         .217 

Original  Exercises  (Theorems) 230 

Constructions       ..........  233 

Formulas      ...........  236 

Original  Exercises  (Numerical) 239 

Original  Constructions 244 

Maxima  and  Minima  .         .         .         .        .        .        .         .         .  245 

Original  Exercises 249 

INDEX  OF  DEFINITIONS     .........  251 


PLANE   GEOMETRY 


INTRODUCTION 

1.  Geometry  is  a  science  which  treats  of  the  measurement 
of  magnitudes. 

2.  A  definition  is  a  statement  explaining  the  significance 
of  a  word  or  a  phrase. 

Every  definition  should  be  clear,  simple,  descriptive,  and  correct ;  that 
is,  it  should  contain  the  essential  qualities  or  exclude  all  others,  or  both. 

3.  A  point  is  that  which  has  position  but  not  magnitude. 

4.  A  line  is  that  which  has  length  but  no  other  magnitude. 

5.  A  straight  line  is  a  line  which  is  determined  (fixed  in 
position)  by  any  two  of  its  points.     That  is,  two  lines  that 
coincide  entirely,  if  they  coincide  at  any  two   points,  are 
straight  lines. 

6.  A  rectilinear  figure  is  a  figure  containing  straight  lines 
and  no  others. 

7.  A  surface  is  that  which  has  length  and  breadth  but  no 
other  magnitude. 

8.  A  plane  is  a  surface  in  which  if  any  two  points  are 
taken,  the  straight  line  connecting  them  lies  wholly  in  that 
surface. 

9.  Plane  Geometry  is  a  science  which  treats  of  the  proper- 
ties of  magnitudes  in  a  plane. 

10.   A  solid  is  that  which  has  length,  breadth,  and  thick- 
ness.    A  solid  is  that  which  occupies  space. 

11 


12  PLANE   GEOMETRY 

11.  Boundaries.     The  boundaries  (or  boundary)  of  a  solid 
are  surfaces.     The   boundaries  (or  boundary)  of  a  surface 
are    lines.     The    boundaries    of   a   line    are    points.     These 
boundaries  can  be  no  part  of  the  things  they  limit.     A  sur- 
face is  no  part  of  a  solid  ;  a  line  is  no  part  of  a  surface  ;  a 
point  is  no  part  of  a  line. 

12.  Motion.     If  a  point  moves,  its  path  is  a  line.     Hence, 
if  a  point  moves,  it  generates  (describes  or  traces)  a  line ;  if 
a  line  moves  (except  upon   itself),  it  generates   a  surface  ; 
if  a  surface  moves  (except  upon  itself),  it  generates  a  solid. 

NOTE.     Unless  otherwise  specified  the  word  "  line  "  hereafter  means 
straight  line. 

ANGLES 


ANGLE  ADJACENT     VERTICAL  ANGLES      RIGHT  ANGLES 

ANGLES  PERPENDICULAR 

13.  A  plane  angle  is  the  amount  of  divergence  of  two 
straight   lines   that   meet.      The  lines  are    called    tjie    sides 
of  the  angle.     The  vertex  of  an  angle  is  the  point  at  which 
the  lines  meet. 

14.  Adjacent   angles  are  two  angles  that  have  the  same 
vertex  and  a  common  side  between  them. 

15.  Vertical    angles    are  two  angles   that  have   the  same 
vertex,  the  sides  of  one  being  prolongations  of  the  sides  of 
the  other. 

16.  If  one  straight  line  meets  another  and  makes  the  ad- 
jacent angles  equal,  the  angles  are  right  angles. 


INTRODUCTION  13 

17.  One  line  is  perpendicular  to  another  if  they  meet  at 
right  angles.     Either  line  is  perpendicular  to  the  other.     The 
point  at  which  the  lines  meet  is  the  foot  of  the  perpendicular. 
Oblique  lines  are  lines  that  meet  but  are  not  perpendicular. 

18.  A  straight  angle  is  an  angle  whose  sides  lie  in  the 
same  straight  line,  but  extend  in  opposite   directions  from 
the  vertex. 


OBTUSE  ANGLE        ACUTE     COM  PLEMENTARY  SUPPLEMENTARY  ANGLES 
ANGLE         ANGLES 

19.  Aii  obtuse  angle  is  an  angle  that  is  greater   than   a 
right  angle.     An  acute  angle  is  an  angle  that  is  less  than 
a    right    angle.     An    oblique    angle   is    any    angle   that   is 
not  a  right  angle. 

20.  Two  angles  are  complementary  if  their  sum  is  equal  to 
one  right  angle.     Two  angles  are  supplementary  if  their  sum 
is  equal  to  two  right  angles.     Thus  the  complement  of  an 
angle  is  the  difference  between  one  right  angle  and  the  given 
angle.    The  supplement  of  an  angle  is  the  difference  between 
two  right  angles  and  the  given  angle. 

21.  A  degree    is  one   ninetieth    of   a   right   angle.     The 
degree  is  the  familiar  unit  used  in  measuring  angles.     It  is 
evident  that  there  are   90°  in  a  right  angle ;    180°  in  two 
right  angles,  or  a  straight  angle ;  360°  in  four  right  angles. 

22.  Notation.    A  point  is  usually  denoted  by  a  capital  letter,  placed 
near  it.     A  line  is  denoted  by  two  capital  letters,  placed  one  at  each  end, 
or  one  at  each  of  two  of  its  points.     Its  length  is  sometimes  represented 
advantageously  by  a  small  letter  written  near  it.     Thus,  the  line  AB; 
the  line  RS',  the  line  m. 

A B   5 §_       ™ : 


14 


PLANE   GEOMETRY 


An  angle  is  usually  denoted  by  three  capital  letters,  placed  one  at  the 
vertex  and  one  on  each  side.  If  only  one  angle  is  at  a  vertex,  the  capital 
letter  at  the  vertex  is  sufficient  to  designate  the  angle.  Sometimes  it  is 
advantageous  to  name  an  angle  by  a  small  letter  placed  within  the  angle. 
The  word  "  angle  "  is  usually  denoted  by  the  symbol  "  Z  "  in  geometrical 


x  o 

Z  a  AND  Z 
Z  XMA  OR  Z  M.          NOT  Z  0 

It  is  important  that  in  naming  an  angle  by  the  use  of  three  letters,  the 
vertex-letter  should  be  placed  between  the  others.  The  size  of  an  angle 
does  not  depend  upon  the  length  of  the  sides,  but  only  on  the  amount  of 
their  divergence.  Thus,  Z  x  =  Z  P  and  Z  P  is  the  same  as  Z  APR  or 
Z  APS  or  Z  BPS,  etc.  An  angle  is  said  to  be  included  by  its  sides.  An 
angle  is  bisected  by  a  line  drawn  through  the  vertex  and  dividing  the 
angle  into  two  equal  angles. 

TRIANGLES 

23.  A  triangle  is  a  portion  of  a  plane  bounded  by  three 
straight  lines.  These  lines  are  the  sides.  The  vertices  of  a 
triangle  are  the  three  points  at  which  the  sides  intersect. 
The  angles  of  a  triangle  are  the  three  angles  at  the  three 
vertices.  Each  side  of  a  triangle  has  two  angles  adjoining 
it.  The  symbol  for  triangle  is  "A". 

A 


ISOSCELES  A 


RIGHT  A         OBTUSE  A  ACUTE  A 

SCALENE  & 


EQUILATERAL  A 
EQUIANGULAR  A 

The  base  of  a  triangle  is  the  side  on  which  the  figure  appears 
to  stand.  The  vertex  of  a  triangle  is  the  vertex  opposite 
the  .base.  The  vertex-angle  is  the  angle  opposite  the  base. 


15 

24.  Kinds  of  triangles  : 

A  scalene  triangle  is  a  triangle  no  two  sides  of  which  are  equal. 

An  isosceles  triangle  is  a  triangle  two  sides  of  which  are  equal. 

An  equilateral  triangle  is  a  triangle  all  sides  of  which  are  equal. 

A  right  triangle  is  a  triangle  one  angle  of  which  is  a  right  angle. 

An  obtuse  triangle  is  a  triangle  one  angle  of  which  is  an  obtuse  angle. 

An  acute  triangle  is  a  triangle  all  angles  of  which  are  acute  angles. 

An  equiangular  triangle  is  a  triangle  all  angles  of  which  are  equal. 

25.  The  hypotenuse  of  a  right  triangle  is  the   side  op- 
posite the  right  angle.     The  sides  forming  the  right  angle 
are  called  the  legs.     In  an  isosceles  triangle  the  equal  sides 
are  sometimes  called  the  legs,  and  the  other  side,  the  base. 

26.  Homologous  Parts.      If  two  triangles  have  the  three 
angles  of  one  equal  respectively  to  the  three  angles  of  the 
other,  the  pairs  of  equal  angles  are  homologous.     Homologous 
sides  in  two  triangles  are  opposite  the  homologous  angles. 

27.  Homologous  parts  of  equal  figures  are  equal. 

If  the  triangles  DEF  and 
HIJ  are  equal  in  all  respects, 
Z  D  is  homologous  to,  and 
=  Z  H,  hence  EF  is  homolo- 
gous to,  and  =  U.  And 
Z  E  is  homologous  to,  and  =  Z  /,  hence,  DF  is  homologous 
to,  and  =  flj,  and  so  on. 

SUPERPOSITION.    SYMBOLS 

28.  Equality  and  coincidence.      Two   geometrical   figures 
are  equal  if  they  can  be  made  to  coincide  in  all  respects. 
Angles  coincide,  and  are  equal,  if  their  vertices  are  the  same 
point  and  the  sides  of  one  angle  are  identical  with  the  sides 
of  the  other.     Superposition  is  the  process  of  placing  one 
figure  upon  another.     This  method  of  showing  the  equality 
of  two  geometrical  figures  is  employed  only  in  establishing 
fundamental  principles. 


16 


PLANE   GEOMETRY 


29.    Symbols.    The   usual  symbols  and  abbreviations  em- 
ployed in  geometry  are  the  following: 


4-  plus. 
-  minus. 

=  equals,  or   is    (or  are) 
equal  to. 

—  is  (or  are)  equivalent  to. 

>  is  (or  are)  greater  than. 

<  is  (or  are)  less  than. 

.'.  hence,  therefore,  conse- 
quently. 

J_  perpendicular. 

Js  perpendiculars. 

O  circle. 


<D  circles. 
Z  angle. 
A  angles. 

rt.  Z  right  angle, 
rt.  A  right  angles. 
A  triangle. 
A  triangles, 
rt.  A  right  triangles. 
||  parallel. 
||s  parallels. 
O  parallelogram. 
HJ  parallelograms. 


ax.  axiom, 

hyp.  hypothesis, 

comp.  complementary. 

supp.  supplementary, 

const,  construction, 

cor.  corollary, 

st.  straight, 

def.  definition, 

alt.  alternate, 

int.  interior, 

ext.  exterior. 


AXIOM,    POSTULATE,    AND    THEOREM 

30.  An  axiom  is  a  truth  assumed  to  be  self-evident.     It  is 
a  truth  which  is  received  and  assented  to  immediately. 

31.  AXIOMS. 

1.  Magnitudes  that  are  equal  to  the  same  thing,  or  to  equals,  are 
equal  to  each  other. 

2 .  If  equals  are  added  to,  or  subtracted  from,  equals,  the  results  are 
equal. 

3.  If  equals  are  multiplied  by,  or  divided  by,  equals,  the  results  are 
equal, 

[Doubles  of  equals  are  equal ;  halves  of  equals  are  equal.] 

4.  The  whole  is  equal  to  the  sum  of  all  of  its  parts. 

5.  The  whole  is  greater  than  any  of  its  parts. 

6.  A  magnitude  may  be  displaced  by  its  equal  in  any  process. 
[Briefly  called  "  substitution."] 

7.  If  equals  are  added  to,  or  subtracted  from,  unequals,  the  results 
are  unequal  in  the  same  sense. 

8.  If  unequals  are  added  to  unequals  in  the  same  sense,  the  results 
are  unequal  in  that  sense. 

9.  If  unequals  are  subtracted  from  equals,  the  results  are  unequal 
in  the  opposite  sense. 


INTRODUCTION  17 

10.  Doubles  or  halves  of  unequals  are  unequal  in  the  same  sense. 

11.  If  the  first  of  three  magnitudes  is  greater  than  the  second,  and 
the  second  is  greater  than  the  third,  the  first  is  greater  than  the  third. 

12.  A  straight  line  is  the  shortest  line  that  can  be  drawn  between 
two  points. 

13.  A   geometrical   figure   may   be   moved   from    one  position  to 
another  without  any  change  in  form  or  magnitude. 

32.  A  postulate  is  something  required  to  be  done,  the  pos- 
sibility of  which  is  admitted  as  evident. 

33.  POSTULATES. 

1.  It  is  possible  to  draw  a  straight  line  from  any  point  to  any 
other  point. 

2.  It  is  possible  to  extend  (prolong  or  produce)  a  straight  line  in- 
definitely, or  to  terminate  it  at  any  point. 

34.  A  geometric  proof  or.  demonstration  is  a  logical  course 
of  reasoning  by  which  a  truth  becomes  evident. 

35.  A  theorem  is  a  statement  that  requires  proof. 

In  the  case  of  the  preliminary  theorems  which  follow,  the 
proof  is  very  simple  ;  but  as  these  theorems  are  not  self- 
evident  they  cannot  be  classified  with  the  axioms. 

A  corollary  is  a  truth  immediately  evident,  or  readily  estab- 
lished, from  some  other  truth  or  truths. 


EXERCISE  1.    Draw  an  /.ABC.     In  ZABC  draw  line  ED. 

What  does  Z  ABD  +  Z  DBC  =  ? 

What  does  Z  ABC  -  Z  A  BD  =  ? 

Ex.  2.  In  a  rt.  Z  ABC  draw  line  BD. 

If  Z  ABD =25°,  how  many  degrees  are  there  in  ZDBCt 

How  many  degrees  are  there  in  the  complement  of  an  angle  of  38°  ? 
How  many  degrees  are  there  in  the  supplement  ? 

Ex.  3.    Draw  a  straight  line  AB  and  take  a  point  X  on  it. 

What  line  does  AX  +  BX  =  ? 

What  line  does  A  B  -  BX  =  ? 

Ex.  4.  Draw  a  straight  line  AB  and  prolong  it  to  .Y  so  that  BX  =  AB> 
Prolong  it  so  that  AX  =  AB. 

BOBBINS1    PLANE    GEOM.  — 2 


BOOK    I 

ANGLES,  LINES,  RECTILINEAR  FIGURES 
PRELIMINARY   THEORIES 

36.  A  right  angle  is  equal  to  half  a  straight  angle. 
Because  of  the  definition  of  a  right  angle.      (See  16.) 

37.  A  straight  angle  is  equal  to  two  right  angles.     (See  36.) 

38.  Two  straight  lines  can  intersect  in  only  one  point. 
Because  they  would  coincide    entirely  if  they   had  two 

common  points.    (See  5.) 

39.  Only  one  straight  line  can  be  drawn  between  two  points. 
(See  5.) 

40.  A  definite  (limited  or  finite)  straight  line  can  have  only  one 
midpoint. 

Because  the  halves  of  a  line  are  equal. 

41.  All  straight  angles  are  equal. 

Because  they  can  be  made  to  coincide.  (See  28  and  Ax. 
13.) 

42.  All  right  angles  are  equal. 

They  are  halves  of  straight  angles  (36),  and  hence  equal 
(Ax.  3). 

43.  Only  one  perpendicular  to  a  line  can  be  drawn  from  a  point  in 
the  line. 

Because  the  right  angles  would  not  be  equal  if  there 
were  two  perpendiculars  ;  and  all  right  angles  are  equal. 
(See  42.) 

18 


BOOK  I  19 

44.  If  two  adjacent  angles  have  their  exterior  sides  in  a  straight 
line,  they  are  supplementary. 

Because  they  together  =  two 
rt.  A.     (See  20.) 

45.  If  two  adjacent  angles  are 
supplementary,  their  exterior  sides      _ 
are  in  the  same  straight  line. 

Because  their  sum  is  two  rt.  A  (20);  or  a  straight  Z 
(37).  Hence  the  exterior  sides  are  in  the  same  straight 
line  (18). 

46.  The  sum  of  all  the  angles  on  one  side  of  a  straight  line  at  a 
point  equals  two  right  angles. 

(See  Ax.  4  and  37.) 

47.  The  sum  of  all  the  angles  about 
a  point  in  a  plane  is  equal  to  four 
right  angles.     (See  46.) 


48.  Angles  that  have  the  same  complement  are  equal.    Or,  comple- 
ments of  the  same  angle,  or  of  equal  angles,  are  equal. 

Because  equal  angles  subtracted  from  equal  right  angles 
leave  equals.  (See  Ax.  2.) 

49.  Angles  that  have  the  same  supplement  are  equal.    Or,  supple- 
ments of  the  same  angle,  or  of  equal  angles,  are  equal.     (  See  Ax.  2.) 

50.  If  two  angles  are  equal  and  supplementary,  they  are  right 
angles. 

Because  each  is  half  a  straight  Z;  hence  each  is  a  rt.  Z. 
(See  36.) 

NOTE.  A  single  number,  given  as  a  reference,  always  signifies  the 
truth  stated  in  that  paragraph  and  is  usually  the  statement  in  full  face 
type  only.  Tn  reciting  or  writing  the  demonstrations  the  pupil  should 
quote  the  correct  reason  for  each  statement,  and  not  give  the  number 
of  its  paragraph.  [Consult  model  demonstrations  on  page  24.] 


20 


PLANE  GEOMETRY 


THEOREMS  AND  DEMONSTRATIONS 

51.  THEOREM.     Vertical  angles  are  equal. 

Given:  A  AOM  and  BOL,  a 
pair  of  vertical  angles. 

To  Prove :  Z  A OM  =  Z  BOL. 

Proof :  /.AOM  is  the  supple- 
ment of  Z  MOB.  (Why?) 
(See  44.) 

ZBOL  is  the  supplement  of  Z  MOB.     (Why?)    (See  44.) 

.-.  /LAOM=/.BOL.    (Why  ?)  (See  49.) 

A  A  OL  and  BOM  are  a  pair  of  vertical  angles.  These  may  be  proven 
equal  in  precisely  the  same  manner.  If  Z  A  OL  =  80°,  how  many  degrees 
are  there  in  the  other  zt? 

52.  THEOREM.   Two  triangles  are  equal  if  two  sides  and  the  in- 
cluded angle  of  one  are   equal  respectively  to  two  sides  and  the 
included  angle  of  the  other. 

c  T 


Given :     A  ABC&ud'RST;  AB  =  RS\  AC  —  llT\  Z  A  =  Z  tf. 
To  Prove  :     A  ABC  =  A  RST. 

Proof :     Place  the  A  ABC  upon  the  A  RST  so  that  Z  A  co- 
incides with  its  equal,  Z  R ;    then  AB  will  fall  upon  RS  and 
point  B  upon  8.     (It  is  given  that  AB  =  RS.)     AC  will  fall 
upon  RT  and  point  C  upon  T.     (It  is  given  that  AC  =  RT.) 
/.  BC  will  coincide  with  ST.     (Why?)  (See.  39.) 
Hence,  the  triangles  coincide  in  every  respect   and   are 
equal  (28). 


BOOK    I 


21 


53.  THEOREM.    Two  right  triangles  are  equal  if  the  two  legs  of 
one  are  equal  respectively  to  the  two  legs  tf  the  other. 

Given  :  Ht.  A  ABC  and  DEF', 
AC  —  DF\  CK  =  FE. 

To  Prove :   A  ABC  =  A  DEF. 

Proof:  In  the  A  ABC  and 
DEF,  AC  =  DF  (Given)  ;  CB 
=  FE  (Given)  ;  Z  C  =  Z  F. 
(Why?)  (See  42.) 

/.  the  A  are  equal.  (Why?) 
(Theorem  of  52.) 

54.  THEOREM.    Two  triangles  are  equal  if  a  side  and  the  two 
angles  adjoining  it  in  the  one  are  equal  respectively  to  a  side  and 
the  two  angles  adjoining  it  in  the  other. 


Given:    A  BCD  and  JKL  ;     BC  =  JK  ;     Z  B  =  Z  J  ;     Z  C  = 


To  Prove  :    A  BCD  =  A  JKL. 

Proof:    Place  A  BCD  upon  A  JKL  so  that  Z  B  coincides 

with  its  equal,  Z  J,  BC  falling  on  JK. 

Point  C  will  fall  on  JT.     (It  is  given  that  BC=  JK.) 
BD  will  fall  on  JL.     (Because  Z  I?  is  given  =  Z  J.) 
CD  will  fall  on  7TL.     (Because  Z  C  is  given  =  Z  K.) 
Then  point  D  which  falls  on  both  the  lines  JL  and  KL  will 

fall  at  their  intersection,  L.    (Why  ?)     (See  38.) 
.-.  the  A  are  =.     (Why?)     (See  28.) 


22 


PLANK   GEOMETRY 


55.  THEOREM.    The  angles  opposite  the  equal  sides  of  an  isosceles 
triangle  are  equal.  A 

Given  :     A  ABC,  AB  =  AC. 
To  Prove  :     Z  B  =  Z  C. 

Proof:  Suppose  AX  is 
drawn  dividing  Z  BAG  into 
two  equal  angles,  and  meeting 
BC  at  X.  In  A  BAX  and 
CAX,  AX=AX  (Identical); 
AB=AC  (Given);  Z  BAX  B~~  * 

=  Z  (MX.     (Because  AX  made  them= .)    .-.  A  ABX= A 
(Why?)  (52.)     .-.ZB  =  ZC.   (Why?)   (See  27.) 

56.  THEOREM.    An  equilateral  triangle  is  equiangular.   (See  55.) 

57.  THEOREM.    The  line  bisecting  the  vertex-angle  of  an  isosceles 
triangle  is  perpendicular  to  the  base,  and  bisects  the  base. 

Prove  A  ABX  and  ACX  equal  as  in  55.  Then,  Z  AXB 
=  Z  AXC.  (Why?)  (27.)  /.  A  AXB  and  AXGme  rt.  A  (16). 

/.  AX  is  _L  to  BC.  (Why?)  (17.)  And,  also,  BX=  CX. 
(Why?)  (27.) 

58.  THEOREM.    Two  triangles  are  equal,  if  the  three  sides  of  one 
are  equal  respectively  to  the  three  sides  of  the  other. 


Given :    A  ABC  and  RST; 
AC=ET\  BC=  ST. 

To  Prove :   A  RST=  A  ABC. 


BOOK  I  23 

Proof  :     Place  A  ABC  in  the  position  of  A  AST  so  that  the 
longest  equal  sides  (TiC  and  ST)  coincide  and  A  is  opposite 
ST  from  B.     Draw  724.      BS  =  AS  (Given).     .*.  ASB  is  an 
isosceles  A.     (Def.  24.) 
/.  Z  SRA  =  Z  SAB.  (Why?)  (55.)  Likewise  TB  =  AT  (?)  and 

Z  TEA  =  Z  7^7?.  (Why?)    Adding  these  equals  we  obtain 

Z  SET  =  Z  SAT   (Ax.  2).      /.  A  BST  =  A  AST  (52). 

That  is,  A  BST=A  ABC.     (Substitution,  Ax.  6.) 

59.  Elements  of  a  theorem.      Every  theorem  contains  two 
parts,  the  one  is  assumed  to  be  true  and  the  other  results' 
from  this  assumption.     The  one  part  contains  the  given  con- 
ditions, the  other  part  states  the  resulting  truth. 

The  assumed  part  of  a  theorem  is  called  the  hypothesis. 

The  part  whose  truth  is  to  be  proved  is  the  conclusion. 

Usually  the  hypothesis  is  a  clause  introduced  by  the  word 
"if."  When  this  conjunction  is  omitted,  the  subject  of  the 
sentence  is  known  and  its  qualities,  described  in  the  quali- 
fying words,  constitute  the  "given  conditions."  Thus,  in 
the  theorem  of  58,  the  assumed  part  follows  the  word  "  if," 
and  the  truth  to  be  proved  is:  "Two  triangles  are  equal." 

60.  Elements  of  a  demonstration.     All  correct  demonstra- 
tions should  consist  of  certain  distinct  parts,  namely : 

1.  Full  statement  of  the  given  conditions  as  applied  to  a 
particular  figure. 

2.  Full  statement  of  the  truth  which  it  is  required  to  prove. 

3.  The  Proof.     This  consists  in  a  series  of  successive  state- 
ments, for  each  of  which  a  valid  reason  should  be  quoted. 
(The  drawing  of  auxiliary  lines  is  sometimes  essential,  but 
this    part  is    accomplished    by    imperatives    for    which    no 
reasons  are  necessary.) 

4.  The  conclusion  declared  to  be  true. 

The  letters  "Q.E.D."  are  often  annexed  at  the  end  of  a 
demonstration  and  stand  for  "  quod  erat  demonstrandum" 
which  means,  "  which  was  to  be  proved." 


24 


PLANE   GEOMETRY 


MODEL    DEMONSTRATIONS 


The  angles  opposite  the  equal  sides  of  an  isosceles  triangle  are  equal. 

Given  :   A  ABC;  AB  =  A  C. 

To  Prove  :  Z  B  =  Z  C. 

Proof  :     Suppose    A  X     is     drawn    bisecting 


B 


^.BAC  and  meeting  BC  at  X. 
In  the  &  BA  X  and  C^Y 
,4Z  =  .4A'  (Identical). 
AB  =  AC  (Hypothesis). 
/.BAX  =  Z  CAX  (Construction). 

.'.&ABX  =  &ACX.  (Two  &  are  =  if  two  sides  and  the  included  Z  of 
one  are  =  respectively  to  two  sides  and  the  included  Z  of  the  other.) 
Hence,  Z  B  =  Z  C.  (Homologous  parts  of  equal  figures  are  equal.)  Q.E.D. 


Two  triangles  are  equal  if  the  three  sides  of  one  are  equal  respectively  to 
the  three  sides  of  the  other. 

Given  :  A  ABC  and 
RST;  AB  =  RS;  AC= 
RT;  BC=ST. 

To  Prove  :  A  RST  = 
A  ABC. 

Proof  :  Place  A  ABC  in 
the  position  of  A  A  ST  so 
that  the  longest  equal 
sides  (BC  and  ST)  coincide,  and  A  is  opposite  5 T7 from  R.  Draw  RA. 

RS  =  AS  (Hypothesis). 

A  A  SR  is  isosceles.  (An  isosceles  A  is  a  A  two  sides  of  which  are  equal.) 
.'.  Z  SRA  =  ZSAR...(\)..(TheA  opp.  the  =  sides  of  an  isos.  A  are  = .) 

Again,  TR  =  A  T  (Hypothesis). 
A  TRA  is  isosceles.  (Same  reason  as  before.) 
(Same  reason  as  for  (1).)  Adding  equations 

(1)  and  (2). 

Z  SRT  =  Z  SA  T.  (If  =  's  are  added  to  =  's  the  results  are  = .) 
Consequently,  the  A  RST  =  A  AST.     (Two  A  are  =  if  two  sides  and 
the  included  Z  of  one  are  =  respectively  to  two  sides  and  the  included 
Z.  of  the  other.) 

That  is,  A  RST  =  A  ABC.     (Substitution  ;    A  ABC  is  the  same   as 
&AST.)  Q.E.D. 


TRA  =  Z  TA  R . . .  (2) 


BOOK    I  25 

The  preceding  form  of  demonstration  will  serve  to  illustrate  an  excel- 
lent scheme  of  writing  the  proofs.  It  will  be  observed  that  the  statements 
are  at  the  left  of  the  page  and  their  reasons  at  the  right.  This  arrange- 
ment will  be  found  of  great  value  in  the  saving  of  time,  both  for  the 
pupil  who  writes  the  proofs  and  for  the  teacher  who  reads  them. 

61.  The  converse  of  a  theorem  is  the  theorem  obtained  by 
interchanging  the  hypothesis  and  conclusion  of  the  original 
theorem.     Consult  44  and  45;   79,  80,  and  others. 

Every  theorem  which  has  a  simple  hypothesis  and  a  simple 
conclusion  has  a  converse,  but  only  a  few  of  these  converses 
are  actually  true  theorems. 

For  example  :  Direct  theorem  :  "  Vertical  angles  are 
equal." 

Converse  theorem  :  "  If  angles  are  equal,  they  are  verti- 
cal." This  statement  cannot  be  universally  true. 

The  theorem  of  120  is  the  converse  of  that  of  55. 

62.  Auxiliary  lines.    Often  it  is  impossible  to  give  a  simple 
demonstration  without -drawing  a  line  (or  lines)  not  described 
in  the  hypothesis.    Such  lines  are  usually  dotted  for  no  other 
reason  than  to  aid   the  learner  in  distinguishing   the  lines 
mentioned  in  the  hypothesis  and  conclusion  from  lines  whose 
use  is  confined  to  the  proof.     Hence,  lines  mentioned  in  the 
hypothesis  and  conclusion  should   never  be   dotted.    (The 
figure  used  in  57  should  contain  no  dotted  line.) 

63.  Superposition.    It  is  worthy  of  note  that  demonstration 
by  the  method  of  superposition  is  never  employed  unless  the 
hypothesis  gives  a  pair  of  equal  angles. 

64.  Homologous  parts.    Triangles  are  proven  equal  in  order 
that  their  homologous  sides,  or  homologous  angles,  may  be 
proven  equal.     This  is  a  very  common  method  of  proving 
lines  equal  and  angles  equal. 

65.  The  distance  from  one  point  to  another  is  the  length 
of  the  straight  line  joining  the  two  points. 


26 


PLANE   GEOMETRY 


66.   THEOREM.    If  lines  be  drawn  from  any  point  in  a  perpendicu- 
lar erected  at  the  midpoint  of  a  straight  line  to  the  ends  of  the  line, 
I.    They  will  be  equal. 

II.    They  will  make  equal  angles  with  the  perpendicular. 
III.   They  will  make  equal  angles  with  the  line. 

Given  :  AB  _i_  to  CD  at  its 
midpoint,  B ;  P  any  point  in 
AB  ;  PC  and  PD. 

To  Prove  :  I.  PC=PD  ; 
II.  Z  CPB  =  Z  DPB  ;   and 
III.  Z  C  =  Z  D. 

Proof:    In    rt.  A    PBC   and    c 
PBD,  BC  =  BD  (Hyp.)  ;  BP  =  BP  (Iden.). 


=  Z  DP.B  (  Why  ?)  ; 
Q.E.D. 


.    (Why?)    (53.) 

/.    I.    PC=PD(Why?)  (27;)  II.  Z 
III.  Zc  =  ZD   (Why?). 


67.  THEOREM.  Any  point  in  the  perpendicular  bisector  of  a  line  is 
equally  distant  from  the  extremities  of  the  line.     (See  66,  I.) 

68.  THEOREM.     Any  point  not  in  the  perpendicular  bisector  of  a 
line  is  not  equally  distant  from  the  extremities  of  the  line. 

Given  :  AB  _L  bisector  of  CD  ; 
P  any  point  not  in  AB  ;  PC 
and  PD.  o 

To  Prove  :  PC  not  =  PD. 

Proof  :  Either  PC  or  PD  will 
cut  AB. 

Suppose  PC  cuts  AB  at  O. 
Draw  OD. 

DO  +  OP  >  PD.    (Why  ?)  (Ax.  12.)    But  CO  =  OD  (67). 

.'.  co+  OP  >  PD.  (Substitution;  Ax.  6.) 

That  is,  PC  >  PD,  or  PC  is  not  =  PD.  Q.E.D. 


BOOK    I 


27 


69.  THEOREM.   If  a  point  is  equally  distant  from  the  extremities  of 
a  line,  it  is  in  the  perpendicular  bisector  of  the  line.  (See  67  and  68.) 

70.  THEOREM.   Two  points  each  equally  distant  from  the  extrem- 
ities of  a  line  determine  the  perpendicular  bisector  of  the  line. 

Each  point  is  in  the  _L  bisector  (69);   two  points  deter- 
mine a  line  (5). 

71.  THEOREM.    Only  one  perpendicular  can  be  drawn  to  a  line  from 
an  external  point. 

Given:  P7?  J_  to  AB  from  P; 
PD  any  other  line  from  P  to  AB. 

To  Prove :  PD  cannot  be  J_  to 
AB ;  that  is,  PR  is  the  only  J_  to 
AB  from  P. 

Proof:  Extend  PR  to  s,  mak- 
ing RS  =  PR  ;  draw  DS. 

In  rt.  A  PDR  and  SDR,  PR  =  RS 
(Const.). 

DR  =  DR  (Iden.).  .'.A  PDR  = 
A  SDR.  (Why?)  (53.) 

.-.  Z  PDR  =  Z  SDR  (27).    That  is,  Z  PDR  =  half  of  Z  PDS. 

Now  PRS  is  a  straight  line  (Const.). 

.'.  PDS  is  not  a  straight  line  (39). 

.-.  Z  PDS  is  not  a  straight  angle  (18). 

.-.  Z  PDR,  the  half  of  Z  PDS,  is  not  a  right  angle  (36). 

.-.  PD  is  not  _L  (17).     .'.  PR  is  the  only  J_.  Q.E.D. 


Ex.  1.  Through  how  many  degrees  does  the  minute  hand  of  a  clock 
move  in  15  min.  ?  in  20  min.  ?  Through  how  many  degrees  does  the  hour 
hand  move  in  one  hour?  in  45  minutes?  in  10  minutes? 

Ex.  2.  How  many  degrees  are  there  in  the  angle  between  the  hands 
of  a  clock  at  9  o'clock?  at  10  o'clock?  at  12  :30?  at  2  : 15  ?  at  3  : 45? 

Ex.  3.  THEOREM.  If  two  lines  be  drawn  bisecting  each  other,  and 
their  ends  be  joined  in  order,  the  opposite  pairs  of  triangles  will  be 
equal.  [Use  51  and  52.] 


PLANE   GEOMETRY 


72.  THEOREM.  Two  right  triangles  are  equal  if  the  hypotenuse 
and  an  adjoining  angle  of  one  are  equal  respectively  to  the  hypote- 
nuse and  an  adjoining  angle  of  the  other. 

N  r 


L  MR  S 

Given:      Rt.  A  LMN  and  RST  ;  LN  —  RT  ;  and  Z  L  =  Zft. 
To  Prove:     A  LMN  =  A  RST. 

Proof  :  Superpose  A  LMN  upon  A  RST  so  that  Z.  L  coincides 
with  its  equal,  Z.  R,  LM  falling  along  RS.  Then  LN  will  fall 
on  RT  and  point  N  will  fall  exactly  on  T  (LN=  RT  by  Hyp.). 

Now  NM  and  TS  will  both  be  _L  to  RS   from    T  (Rt.  A 
by  Hyp.).     /.  NM  will  coincide  with  TS  (71). 
..-.  A  LMN  =  A/?.sr(28).  Q.E.D. 

73.  THEOREM.  Two  right  triangles  are  equal  if  the  hypotenuse 
and  a  leg  of  one  are  equal  respectively  to  the  hypotenuse  and  a  leg 
of  the  other. 

K  R 


M 


l"~  — J  X —  L 

Given  :    Rt.  A  UK  and  LMR  ;  KI  =  RM  ;   KJ  =  RL. 
To  Prove :   A  UK  =  A  LMR. 

Proof :  Place  A  UK  in  the  position  of  A  XLE  so  that 
the  equal  sides,  KJ  and  RL,  coincide  and  /  is  at  X,  oppo- 
site RL  from  M. 


BOOK   I 


29 


Now,  ARLM  and  RLX  are  supplementary.    (Why  ?)  (20.) 

.*.  XLM  is  a  str.  line  (45). 

Also,  A  RMX  is  isosceles.  (Why  ?)  (RX  =  RM  by  Hyp.) 
.-.Z  x=  Z3/.  (Why?)  (55.) 

.-.  A  XLR  =  A  MLR.  (Why?)  (72.)  That  is,  A  UK  = 
ALMR.  (Ax.  (J.)  Q.E.D. 

COR.  The  perpendicular  from  the  vertex  of  an  isosceles  triangle  to 
the  base  bisects  the  base. 

Proof  :   A  XLR  =  A  MLR.     (Why  V)      . .  XL  =  ML.     (Why  ?) 

74.  THEOREM  Two  right  triangles  are  equal  if  a  leg  and  the  ad- 
joining acute  angle  of  one  are  equal  respectively  to  a  leg  and  the 
adjoining  acute  angle  of  the  other. 


A  CD 

.  Given  :    lit.  A  ABC  and  DEF  ;  AC  =  DF ;   Z^  =  Z  D. 
To  Prove  :    A  AEC  =  A  DEF. 

Proof  :  In  the  A  ABC  and  DEF,  AC  =  DF.  (Why?)  (Hyp.) 
Also  Z.A  =  /LD  (Why?)  and  ZC'  =  ZF.  (Why?)  (42.) 
.'.AABC  =  ADEF.  (Why?)  (54.)  Q.E.D. 


Ex.  1.  How  many  pairs  of  equal  parts  must  two  triangles  have,  in 
order  that  they  may  be  proven  equal?  How  many  pairs  is  it  necessary 
to  mention  in  the  case  of  two  right  triangles? 

Ex.  2.  THEOREM.  If  a  perpendicular  be  erected  at  any  point  in  the 
bisector  of  an  angle,  two  equal  right  triangles  will  be  formed.  [Use  74.] 

Ex.   3.   Through  the  midpoint  of  a  line  AB  any  oblique  line  is  drawn  : 

I.    The  lines  JL  to  it  from  A  and  B  are  equal.     [Use  72.] 
II.    The  lines  _L  to  AB  at  A  and  B,  terminated  by  the  oblique  line, 
are  equal.     [Use  74.] 


30 


PLANE   GEOMETRY 


75.  THEOREM.  The  sum  of  two  sides  of  a  triangle  is  greater  than 
the  sum  of  two  lines  drawn  to  the  extremities  of  the  third  side,  from 
any  point  within  the  triangle.  B 

Given :    P,  any  point  in 
A  ABC  i  lines  PA    and  PC. 
To  Prove :    AB    +  BC  > 
AP  +  PC. 

Proof  :    Extend     AP     to 
meet  BC  at   X.  ~ 

AB  +  BX  >  AP  +  PX.  (Why?)  (Ax.  12.) 

CX  +  PX  >  PC.      (Why  ?)  (Ax.  12.)  Add : 


AB  +  BX  +  CX  +  PX  >  AP  +  PC  +  PX(Ax.  8). 

Subtract  PX  =  PX. 
.'.AB  +  BC  >  AP  +  PC  (Ax.  7). 


Q.E.D. 


76.     THEOREM.     If  from  any  point  in  a  perpendicular  to  a  line 
two  oblique  lines  be  drawn, 

I.    Oblique  lines  cutting  off  equal  distances  from  the  foot  of  the 
perpendicular  will  be  equal. 

II.    Equal  oblique  lines  will  cut  off  equal  distances  (converse). 
III.    Oblique  lines  cutting  off  unequal  distances  will  be  unequal,  and 
that  one  which  cuts  off  the  greater  distance  will  be  the  greater. 


•B      A 


I.    Given  :    CD  -L  to  AB  ;    ND  =  MD  ; 
oblique  lines  PN  and  PM.   [First  figure.] 

To  Prove  :   PN  =  PM. 


BOOK    I 


31 


Proof  :    PD  is  J_  bisector  of  NM  (Hyp.).    .-.  PN  =  PM  (67). 

II.  Given  :   CD  _L  to  AB  ;  PN  =  PM.     [First  figure.] 
To  Prove  :   ND  =  MD. 

Proof:    In   the  rt.  A   PND   and  PMD,  PD  =  PD  (Iden.); 

and  PN  =  PM  (Hyp.).    .'.  A  PND  =  A  PMD.   (Why?)  (73.) 

.-.ND=MD.     (Why?)     (27.)  Q.E.D. 

III.  Given :    CD  _L  to  AB ;    oblique   lines   P.R,    PT ;  also 
RD  >  DT.     [Second  figure.] 

To  Prove:   PR  >  PT. 

Proof  :  Because  RD  is  >  DT,  we  may  mark  DS  (on  RD)  = 
DT.  Draw  P8.  Extend  PD  to  X,  making  DX  =  PD ; 
draw  RX  and  SX.  AD  is  _L  to  PX  at  its  midpoint  (Const.). 
.'.PR  =  RX  and  PS  =  sx  (66). 

Now  P.R  +  RX  >  PS  +  sx  (75). 

Hence  P#  +  PR  >  PS  +  PS  (Ax.  6). 

That  is,  2  PR  >  2  PS.    .'.  PR>  PS  (Ax.  10). 

But  P5=  PT  (76,  I).     .'.  P£  >  PT  (Ax.  6).  Q.E.D. 

COR.  Therefore,  from  an  external  point  it  is  not  possible  to  draw 
three  equal  lines  to  a  given  straight  line. 

77.  THEOREM.  The  perpendicular  is  the  shortest  line  that  can  be 
drawn  from  a  point  to  a  straight  line. 

Given:  PR±toAB;  PC  not  _L. 
To  Prove :  Ptf  <  PC. 

Proof :  Extend  PR  to  X,  mak- 
ing RX=PR.  Draw  CX.  PR  + 
RX  <  PC+  CX  (Ax.  12). 

But  AR  is  _L  to  PXat  its  mid-     A~ 
point  (Const.). 

.-.  PC=CX  (66). 

.'.PR+  PR  <  PC+PC  (Ax.  6). 

That  is,  2 PR  <  2 PC. 

/.  PR<  PC  (Ax.  10).    Q.E.D. 


32  PLANE   GEOMETRY 

78.  The  distance  from  a  point  to  a  line  is  the  length  of  the 
perpendicular  from  the  point  to  the  line.     Thus  "  distance 
from  a  line  "  involves  the  perpendicular.     If  the  perpendicu- 
lars from  a  point  to  two  lines  are  equal,  the  point  is  said  to  be 
equally  distant  from  the  lines. 

79.  THEOREM.    Every  point  in  the  bisector  of  an  angle  is  equally 
distant  from  the  sides  of  the  angle. 

Given:  Z  ACE;  bisector  CQ; 
point  P  in  CQ;  distances  PB  and 
PD. 

To  Prove  :    Pit  =  PD. 

Proof  :  A  PBC  and  PDC  are 
rt.  A  (78). 

In  rt.  A  PBC  and  PDC,  PC  =PC  (Iden.)  ;    Z  PCS  =  Z  PCD 
(Hyp.).    /.A  PJ3C=  APDC(?)(72).    .-.  PB  =  PD  (?).  Q.E.D. 

80.  THEOREM.    Every  point  equally  distant  from  the  sides  of  an 
angle  is  in  the  bisector  of  the  angle. 

Given  :    Z  ACE;  P  a  point,  such  that  PB  =  PD  (distances); 
CQ  a  line  from  vertex  of  the  angle,  and  containing  P. 

To  Prove  :    Z  ACQ  =  Z  ECQ. 

Proof:    A  PJ5C  and  PDC  are  right  A  (?).      In  rt.  A  PBC 
and  PDC,  PC  =  PC  (?)  ;   PB  ==  PD  (?). 

?)  (73).       /.Z^CQ  =  Z^CQ(?).    Q.E.D. 


81.  THEOREM.    Any  point  not  in  the  bisector  of  an  angle  is  not 
equally  distant  from  the  sides  of  the  angle.     [Because   if   it  were 
equally  distant,  it  would  be  in  the  bisector    (80).] 

82.  THEOREM.    The  vertex  of  an  angle  and  a  point  equally  distant 
from  its  sides  determine  the  bisector  of  the  angle.     (See  80  and  5.) 


Ex.    Describe  the  path  of  a  moving  point  which  shall  be  equally  dis- 
tant from  two  intersecting  lines. 


BOOK   I  33 

83.  The  altitude  of  a  triangle  is  the  perpendicular  from 
any  vertex  to   the    opposite  side  (prolonged  if  necessary). 
A  triangle  has  three  altitudes.     The  bisec- 
tor of  an  angle  of  a  triangle  is  the   line 

dividing   any    angle    into    equal    angles. 
A  triangle  has  three  bisectors  of  its  an- 

„,,  ,,          .     .  i        •       ,1         T  THE    THREE    MEDIANS 

gles.     1  he  median  ot  a  triangle  is  the  line 

drawn  from  any  vertex  to  the  midpoint  of  the  opposite  side. 

A  triangle  has  three  medians. 

84.  THEOREM.    The  bisectors  of  the  angles  of  a  triangle  meet  in  a 
point  which  is  equally  distant  from  the  sides. 

Given:  A  ABC,  AX  bisect- 
ing Z  A,  B  Y  and  CZ  the  other 
bisectors. 

To  Prove:   AX,     BY,     cz 

meet  in  a  point  equally  dis- 
tant from  AB,  AC,  and  BC. 

Proof:  Suppose  that  AX 
and  BY  intersect  at  O. 

O  in  AX  is  equally  distant  from  AB  and  AC   (?)  (79). 

O  in  BY  is  equally  distant  from  AB  and  BC  (?). 

.-.  point  O  is  equally  distant  from  AC  and  BC   (Ax.  1).* 

.-.  O  is  in  bisector  CZ  (?)  (80). 

That  is,  all  three  bisectors  meet  at  O,  and  O  is  equally  dis- 
tant from  the  three  sides.  Q.E.D. 
*  The  J_  distances  from  O  to  the  three  sides  are  the  three  equals. 


Ex.  1.   Draw  the  three  altitudes  of  an  acute  triangle;  of  an  obtuse 
triangle. 

Ex.  2.   Prove  that  in  an  equilateral  triangle : 

(a)    An  altitude  is  also  a  median.     [Use  73.] 

(6)    A  median  is  also  an  altitude.     [Use  58,  27,  16,  17.] 

(c)  An  altitude  is  also  the  bisector  of  an  angle  of  the  triangle. 

(d)  The  bisector  of  an  angle  is  also  an  altitude.     [Use  52.] 

(e)  The  bisector  of  an  angle  is  also  a  median. 
BOBBINS'  PLANE  GEOM. — 3 


PLANE   GEOMETRY 


85.  THEOREM.    The  three  perpendicular  bisectors  of  the  sides  of  a 
triangle  meet  in  a  point  which  is  equally  distant  from  the  vertices. 

Given  :  A  ABC;  LR,  MS,  NT, 
the  three  J_  bisectors. 

To  Prove :  LR,  MS,  NT  meet 
at  a  point  equally  distant  from 
A  and  B  and  C. 

Proof :  Suppose  that  LR  and 
MS  intersect  at  O. 

O  in  LR  is  equally  distant 
from^  and  5.  (?)  (67.) 

O  in  MS  is  equally  distant 
from  A  and  C.  (?)  ~~N~" 

.-.  point  O  is  equally  distant  from  B  and  C  (Ax.  1).* 

Hence  O  is  in  J_  bisector  NT  (?)  (69). 

That  is,  all  three  _L  bisectors  meet  at  O,  and  O  is  equally 
distant  from  A  and  B  and  C.  Q.E.D. 

*  The  three  lines  from  0  to  the  vertices  are  the  three  equals. 

86.  THEOREM.    If  two  triangles  have  two  sides  of  one  equal  to  two 
sides  of  the  other,  but  the  included  angle  in  the  first  greater  than  the 
included  angle  in  the  second,  the  third  side  of  the  first  is  greater  than 
the  third  side  of  the  second. 


F  H 

Given:   A  ABC,  DEF\    AB  =  DE;  BC=EF;  Z  ABC  >  Z  E. 


BOOK  1  35 

To  Prove :    AC  >  DF. 

Proof :  Superpose  A  DEF  upon  A  ABC  so  that  line  DE 
coincides  with  its  equal  AB,  A  DEF  taking  the  position  of 
A  ABH.  There  will  remain  an  angle,  HBC.  (Z  ABC  is  >  Z  E. ) 

Suppose  EX  to  be  the  bisector  of  Z  HBC,  meeting  AC  at  X. 
Draw  XH. 

In  A  HEX  and  CBX,  EX  =  EX  (?) ;  EH  =  EC  (?) ;  Z  #J3X  = 
ZCJBX(?)  (Const.)-  .'.&UBX  =  A  CBX  (?)  (52).  .-.#x= 
XC  (?). 

Now,  AX  +  XH  >  ^iff  (?).    .-.AX  +  XC  >  ^4fl  (Ax.  6). 

That  is,  JIC  >  IXF.  Q.E.D. 

87.  THEOREM.  If  two  triangles  have  two  sides  of  one  equal  to  two 
sides  of  the  other  but  the  third  side  of  the  first  greater  than  the  third 
side  of  the  second,  the  included  angle  of  the  first  is  greater  than  the 
included  angle  of  the  second.  [Converse.] 


s 


T 

Given  :    A  ABC  and  RST;  AB  =  BS  ;   EC  =  ST  ;   uiC>  £T. 
To  Prove:     Z  £  >Zs. 

Proof  :    It  is   evident   that   Z  U  <  Z  8,  or  Z  B  =  Z  S,  or 
>  ZS. 

1.  If  ZB  <  ZS,  AC  <  ET  (86). 

But  AC  >  #r  (Hyp.).    .\ZB  is  not  <  Z  5. 

2.  If  Z  u  =  Z  8,  the  A  are  =  (52),  and  AC  is  =  Rr  (27). 
But  AC  >  ET  (Hyp.).     /.Z  E  is  wof  =  Z  S. 

3.  /.  the  only  possibility  is  that  Z  E  >  Zs.  Q.E.D. 


36  PLANE   GEOMETRY 

88.  The  preceding  method  of  demonstration  is  termed  the 
method  of  exclusion.     It    consists    in    making    all    possible 
suppositions,  leaving  the  probable  one  last,  and  then  proving 
all  these  suppositions  impossible,  except  the  last,  which  must 
necessarily  be  true. 

89.  The  method  of  proving  the  individual  steps  is  called 
reductio    ad   absurdum    (reduction  to  an  absurd  or  impos- 
sible conclusion).     This    method    consists    in    assuming   as 
false  the  truth  to  be   proved  and  then    showing  that  this 
assumption    leads    to    a   conclusion  altogether   contrary  to 
known  truth  or  the  given  hypothesis.    (Examine  87.)    This 
is  sometimes  called  the  indirect  method.      The  theorems  of 
93  and  94  are  demonstrated  by  a  single  use  of  this  method. 

90.  THEOREM.    If  two  unequal  oblique  lines  be  drawn  from  any  point 
in  a  perpendicular  to  a  line,  they  will  cut  off  unequal  distances  from  the 
foot  of  the  perpendicular,  and  the  longer  oblique  line  will  cut  off  the  greater 
distance.    [Converse  of  76,  III.]  C 

Given:  CD_LtoAB;  PUand 
PS  oblique  lines  ;  PR  >  PS. 

To  Prove  :   DR  >  DS. 

Proof :    It  is  evident  that 

DR  <  DS,    or    DR  =  DS,    or          ^ _ 

DR  >  DS.  A      R~~  ° 

If  DR  <  DS,  PR  <  PS  (76,  III).    But  PR  >  PS  (Hyp.). 

.-.  DR  is  not  <  DS. 

If  DR  =  DS,  PR  =  PS  (76,  I).     But  PR  >  PS  (Hyp.). 

.'.  DR  is  not  =  DS. 

Therefore  the  only  possibility  is  that  DR  >  DS.      Q.E.D. 

91.  Parallel  lines  are  straight  lines  that  lie  in  the  same  plane 
and  that  never  meet,  however  far  extended  in  either  direction. 

92.  AXIOM.   Only  one  line  can  be  drawn  through  a  point  parallel  to 
a  given  line. 


BOOK  r  37 

93.  THEOREM.  Two  lines  in  the  same  plane  and  perpendicular  to 
the  same  line  are  parallel. 

Given  :  CD  and  EF  in  same 
plane  and  both  J_  to  AB. 

To  Prove :   CD  and  EF  \\  .       E 

Proof :  If  CD  and  EF  were  not  II,  they  would  meet  if  suffi- 
ciently prolonged.  Then  there  would  be  two  lines  from  the 
point  of  meeting  J_  to  AB.  (By  Hyp.  they  are  J_  to  AB.) 

But  this  is  impossible  (?)   (71). 

.-.  CD  and  EF  do  not  meet,  and  are  parallel  (91).    Q.E.D. 

94.  THEOREM.  Two  lines  in  the  same  plane  and  parallel  to  the 
same  line  are  parallel. 

Given :  AB  \\  to  RS  and  CD  ||  to  RS  and  in  the  same  plane. 
To  Prove  :   AB  \\  to  CD.  A e 

Proof :  If  AB  and  CD  were 
not  ||,  they  would  meet  if 
sufficiently  prolonged. 

Then  there  would  be  two  lines  through  the  point  of  meet- 
ing ||  to  the  line  RS.  (By  Hyp.  they  are  ||  to  RS). 

But  this  is  impossible  (92). 

.-.  AB  and  CD  do  not  meet,  and  are  ||  (?)  (91).          Q.E.D. 

95.  THEOREM.  If  a  line  is  perpendicular  to  one  of  two  parallels,  it  is 
perpendicular  to  the  other  also. 


Given :   LM  _L  to  AB  and 
AB  ||  to  CD. 

To  Prove  :  LM  _L  to  CD.       X — 7* 


•B 


Y 


Proof:   Suppose    XT   is 
drawn  through  M  J_  to  LM.     AB  is  ||  to  XT  (?)  (93). 
But  AB  is  ||  to  CD  (Hyp.). 
Now  CD  and  XT  both  contain  M  (Const.). 
.'.CD  and  XT  coincide   (92). 
But  LM  is  _L  to  XT  (?).  That  is,  LM  is  _L  to  CD.          Q.E.D. 


38 


PLANE   GEOMETRY 


19' 


96.  If  one  line  cuts  other  lines,  it  is  called  a  transversal. 
Angles  are  formed  at  the  several  intersections,  and   these 
receive  the  following  names  : 

Interior  A  are  between  the  lines  [b,  c,  e,  h~\. 

Exterior  A  are  without  the  lines  [a,  rf,/,  <?]. 

Alternate  A   are  on   opposite   sides  of  the 
transversal  [b  and  h  ;  c  and  e  ;  a  and  g ;  etc.]. 

Alternate-Interior  A   are  b  and  h ;   and  c 
and  e. 

Alternate-Exterior  A  are  a  and  g ;  d  and/. 

Corresponding  <4  are  a  and  e ;  rf  and  Ti ;  i  and/;  c  and  </. 

Adjoining-Interior  zi  are  c  and  A ;  b  and  e. 

Adjacent-Interior  .4  are  b  and  c ;  e  and  A. 

The  left-hand  transversal  makes  eight  angles  similarly  related.  The 
"primes"  are  used  only  to  designate  angles  which  are  different  from 
those  in  the  right-hand  part  of  the  figure. 

97.  THEOREM.   If  a  transversal  intersects  two  parallels,  the  alter- 
nate interior  angles  are  equal. 

Given :    AB  \\  to  CD ;    transversal  EF  cutting  the  Us  at  H 

E 


and  K. 
To  Prove: 


a  =  /.  i  and 


x  — 


v. 


Proof :  Suppose  through 
Jf,  the  midpoint  of  HJT,  ES  is 
drawn  J_  to  AB.  Then  ES  is 
_L  to  CD  (95).  In  rt.  A  EMH 
and  KMS,  HM  =  KM  (Const. ) ; 
Z  EMH=Z  KMS  (?)  (51).  .'.AEMH  =  A  KMS  (?)  (72). 


Z.  x  is  the  supp.  of  Z  a  (?)  (44)  ; 
is  the  supp.  of  Z  i  (?). 


(49). 


Q.E.D. 


EXERCISE.   If 
there  i 


a  =  70°,  in  the  figure  of  97,  how  many  degrees  are 
v?  Z.AHE1   ^EHB't  Z.CKF1   Z.DKF1 


BOOK  1  39 

96.   THEOREM.   If  a  transversal  intersects  two  parallels,  the  corre- 
sponding angles  are  equal. 

Given  :  AB  II  to  CD ;  trans- 
versal EF  cutting  the  Us  and 
forming  the  8  A. 

To  Prove :  Z  «  =  Z  i ;  Z  c 
=  Z  r  ;  Zo  =  Z  a;  Z  n  =  Z  w.  C —  ^^ 

Proof:      Z«  =  Z«      (51);  / 

Z<?  =  Zm  (?);  Z  w  =  Zr(?).  /.Z  <?=Z  r  (?).    Etc.  Q.E.D. 

99.  THEOREM.    If  a  transversal  intersects  two  parallels,  the  alter- 
nate-exterior angles  are  equal. 

Given:    (?).     To  Prove:    (?).     Proof:    Zc  =  Zr  (?);  and 
Zr  =  Zw  (?).     .'.Zc=Zw  (?).    Etc.  Q.E.D. 

100.  THEOREM.    If  a  transversal  intersects  two  parallels,  the  sum  of 
the  adjoining-interior  angles  equals  two  right  angles. 

Given  :  (?).    To  Prove  :    Z  a  +  Z  r  =  2  rt,  A.     Etc. 

Proof:    Z  a  +  Zm  =  2rt.  Z  (?)  (46).    ButZw  =  Zr(?). 
.-.  Z  a  +  Zr=  2  rt.  A  (Ax.  6).    Etc.  Q.E.D. 

101.  THEOREM.    If  a  transversal  intersects  two  lines  and  the  alter- 
nate-interior angles  are  equal,  the  lines  are  parallel.  [Converse  of  97.] 

Given :    AB    and    CD    two 
lines ;  transversal  EF  cutting 
them  at  H  and  K  respectively ;         A 
Z  a  =  Z  HKD. 

To  Prove :  CD  ||  to  AB. 

IR 

Proof :  Through  K,  suppose 
RS  is  drawn  ||  to  AB.     Then 
(?)    (97).     But 


Z.CL=/.HKD  (Hyp.).  .'./LHKS  =  £HKD  (?).  .-.  KD  and  KS 
coincide:  that  is,  CD  and  RS  are  the  same  line.  .'.  CD  is  || 
to  AH.  (Because  it  coincides  with  RS  which  is  II  to  AB).  Q.E.D. 


40  PLANE   GEOMETRY 

102.   THEOREM.  If  a  transversal  intersects  two  lines  and  the  corre- 
sponding angles  are  equal,  the  lines  are  parallel. 

Given :  AB  and   CD  cut   by 
transversal  EF\  Z  c  —  Z  r. 


To  Prove :  AB  li  to  CD. 

Proof:    Ze  =  Zm  (?);    Z  <? 
=  Zr  (?).    .-.Zw  =  Zr  .(?)•    c 
.'.  AB  IS  II  to  CD  (101).     Q.E.D. 

If  Z  a  were  given  =  Z  o  or 
Z  »  =  Z  i,  etc.,  the  proof  that  AB  is  II  to  CD  would  be  the  same. 

103.  THEOREM.   If  a  transversal  intersects  two  lines  and  the  alter- 
nate-exterior angles  are  equal,  the  lines  are  parallel. 

Given  :  AB  and  CD  cut  by  EF,  and  Z  c  =  Z  n. 
To  Prove :  A  B  II  to  CD. 

Proof:  Zc-Zm  (?);  Z<?=Zn(?).  /.Zm  =  Z/i  (?). 
.'.  ^LBlS  H  to  CD(?).  Etc.  Q.E.D. 

104.  THEOREM.    If  a  transversal  intersects  two  lines  and  the  sum 
of  the  adjoining-interior  angles  equals  two  right  angles,  the  lines  are 
parallel. 

Given  :  AB  and  CD  cut  by  EF  and  Z  a  +  Z  r  =  2  rt.  A 

To  Prove :  ^45  II  to  CD. 

Proof:  Z  «  +  Zc=2  rt.Zs  (46);  Z  a  +  Z  r  =  2  rt,  A  (?). 
.-.Za  +  Z  c=Za  +  Zr  (?).  Hence  Zc  =  Zr  (Ax.  2). 
.-.  .4#  is  II  to  CD  (?)  (102).  Q.E.D. 

r     r  Zaissupp.  of  Z  <?(?);  1  .     ,.>N    ^, 

Another  proof:  J      ....  ."       ;  '      /.Z<?  =  Z  r  (>>  Etc. 

\  Zaissupp.of  Z  r(?).  j 

105.  THEOREM.    If  two  angles  have  their  sides  paraUel  each  to  each, 
the  angles  are  equal  or  supplementary. 

I.    Given :  Z  a  and  Z  b  with  their  sides  II  each  to  each  and 
extending  in  the  same  directions  from  their  vertices. 
To  Prove  :  Z  a  =  Z  b. 
Proof:   If  the  non-parallel  lines  do  not  intersect,  produce 


BOOK   I  41 

them  till  they  meet,  forming  Z  o.    Now, 
Z  a  =  Z  o.  (?)  (98).  Z  o  =  Z  ft  (?). 

/.Z  a  =  Zft  (?).  Q.E.D.     ^ ^ 

II.  Given:   Z  a  and  Z  <?  with  their 

sides  II  each  to  each  and  extending  in  / 

opposite  directions  from  their  vertices.  ' 

To  Prove :  Z  a  =  Z  c. 

Proof  :    Z  a  =  Z  6  (  Proved  in   I ) ; 
Z  5  =  Z  c  (?). 

Hence  Z  a  =  Z  c  (?).  Q.E.D. 

III.  Given :   Z  a  and  Z  c?  with  their  sides  II  each  to  each, 
but  one  pair  extending  in  the  same  direction,  the  other  pair 
extending  in  opposite  directions  from  their  vertices. 

To  Prove  :  Z  a  and  Z  d  supplementary. 
Proof:   Z  b  and  Z  c?  are   supplementary   (44). 
Za  =  Zft(I).  /.Z  aandZ  c?  are  supp.  (Ax.  6.).   Etc.  Q.E.D. 

106.   THEOREM.    If  two  angles  have  their  sides  perpendicular  each  to 
each,  the  angles  are  equal  or  supplementary. 

I .  Given :   A  a  and  b  with 
sides  J_  each  to  each. 

To  Prove  :    Z  a  =  Z  b. 

Proof:  At  B  suppose  BB  is 
drawn  J_  to  BC  and  BS  _L  to  A  B. 
BB  is  II  to  FE  and  BS  is  II  to 
DE  (?)  (93). 

.'.Z  RBS  =  ^b  (?)   (105).  G 

Now,  Z  a  is  the  comp.  of  Z  ABR  (20),  1.-.  Z  a=  Z  UBS  (?). 
and  Z  BBS  is  the  comp.  of  Z  Y!£.R  (?).    J  (48.) 

.'.Za  =  Zft.     (Ax.  1.)  Q.E.D. 

II.  Given :    A  a  and  c  with  sides  _L  each  to  each. 
To  Prove  :    Z  a  and  Z  <? -supplementary. 

Proof:    Z  6  and  Z •<?  are   supp.    (?);    Z  a  =  Z   ft  (I). 

.'.  Z  a  and  Z  c  are  supp.  (Ax.  6).  Q.E.D. 


42  PLANE   GEOMETRY 

107.  An  exterior  angle  of  a  triangle 
is  an  angle  formed  outside  the    tri- 
angle, between  one  side  of  the  tri- 
angle and  another    side    prolonged. 
[Z  ABX.~] 

The   angles   within   the    triangle 
at  the  other  vertices  are  the  opposite  c 
interior  angles.      [Z  A  and  Z  C.] 

108.  THEOREM.    An  exterior  angle  of  a  triangle  is  equal  to  the  sum 
of  the  opposite  interior  angles. 

Given:    A  ABC]    exterior   Z 
ABD. 

To  Prove  :     Z  ABD  =  Z  A  +  B^ p 

Z  C. 

Proof:     Suppose    BP    to    be 
drawn  through  B  II  to  AC. 


Z  ABD  =  Z  ABP  +  Z  P£D  (Ax.  4). 


=  Z  c  (?)  (98). 
/.  Z  ^£1)  =  Z  J.  +  Z  C   (Ax.  6).  Q.E.D. 


109.  THEOREM.   An  exterior  angle  of  a  triangle  is  greater  than 
either  of  the  opposite  interior  angles.       (See  Ax.  5.) 

110.  THEOREM.    The  sum  of  the  angles  of  any  triangle  is  two  right 
angles  ;  that  is,  180°. 

Given  :    A  ABC. 

To  Prove  :    Z  A  -f  Z  B  + 
/.ACS  =  2  rt.  ^  = 


Proof  :    Prolong  AC  to  X, 
making  the  ext.  Z  BOX.          A 


Z  BCX+  Z  ^C£  =  2  rt.  Zs  (?)  (46). 
But    Z  BCX=  Z  ^1  +Z  JS  (?)  (108). 

.-.  Z  ^  +  Z  7?  +  Z  ^<v#  =  -2  rt.  ^  =  180°    (Ax.  6).     Q.E.D. 


BOOK  T  43 

111.  COR.    The  sum  of  any  two  angles  of  a  triangle  is  less  than 
two  right  angles.    (See  Ax.  5.) 

112.  COR.    A  triangle  cannot  have   more  than  one  right  angle  or 
more  than  one  obtuse  angle. 

113.  COR.    Two  angles  of  every  triangle  are  acute.     (See  112.) 

114.  THEOREM.    The  acute  angles  of  a  right  triangle  are  comple- 
mentary. 

Proof :    Their  sum  =  1  rt.  Z    (110  and  Ax.  2).       Hence 
they  are  complementary.     (See  20.) 

115.  COR.    Each  angle  of  an  equiangular  triangle  is  60°. 

116.  THEOREM.    If  two  right  triangles  have  an  acute  angle  of  one 
equal  to  an  acute  angle  of  the  other,  the  remaining   acute   angles 
are  equal.     (See  114  and  48.) 

117.  THEOREM.    If  two  triangles  have  two  angles  of  the  one  equal 
to  two  angles  of  the  other,  the  third  angle  of  the  first  is  equal  to  the 
third  angle  of  the  second.     (See  110  and  Ax.  2.) 

118.  THEOREM.    Two  triangles  are  equal  if  a  side  and  any  two 
angles  of  the  one  are  equal  respectively  to  a  homologous  side  and  the 
two  homologous  angles  of  the  other. 

Proof  :    The  third  Z  of  one  A  =  third  Z  of  other  A  (117). 
/.  the  A  are  =    (54). 

119.  THEOREM.    Two  right  triangles  are  equal  if  a  leg  and  the  op- 
posite acute  angle  of  one  are  equal  respectively  to  a  leg  and  the  opposite 
acute  angle  of  the  other.     (See  118.) 


Ex.  1.  In  the  figure  of  107,  if  Z  A  =  40°  andZ  C  =  70°,  how  many 
degrees  are  there  in  Z  ABXt  in  Z  ABC  ? 

Ex.  2.  If  each  of  the  equal  angles  of  an  isosceles  triangle  is  50°,  how 
many  degrees  are  there  in  the  third  angle  ? 

Ex.  3.  If  one  of  the  acute  angles  of  a  right  triangle  is  25°,  how  many 
degrees  are  there  in  the  other  ? 

Ex.  4.   State  and  prove  the  converse  of  114. 


44 


PLANE   GEOMETRY 


120.  THEOREM.   If  two  angles  of  a  triangle  are  equal,  the  triangle  is 
isosceles.     [Converse  of  55.]  B 

Given:  A  ABC-,  Z^=ZC. 
To  Prove  :  AB  =  EC. 
Proof  :  Suppose  EX  drawn 
_L  to  AC. 

In  rt.  A  A  EX  and  CEX,  EX 

.-.  A  ABX=  A  CBX  (?)    (119).     .'.  AB  =  EC  (?).          Q.E.I). 

121.  THEOREM.   An  equiangular  triangle  is  equilateral. 

122.  THEOREM.   If  two  sides  of  a  triangle  are  unequal,  the  angles 
opposite  them  are  unequal,  and  the  greater  side  subtends  the  greater 
angle. 

Given:    A  ABC',   AB>AC. 
To  Prove  :   Z  ACE  >  Z  B. 

Proof  :    On  AB  take  ^17?  =  AC. 
[We  may,  because  AB  >  AC.'] 

Draw  CR  and  let  Z  ARC=x.    £  B 

Z  ARC  is  an  ext.  Z  of  A  CBR  (?).     .-.  Z  x  >  Z  5  (109). 

Also,  Z  ^iC.R  =  Z  .iflC  =  Z  #  (?)  (55).     Again,  Z  ^4 OB  > 
Z  a;  (?)  (Ax.  5).     .-.  Z  ^c«  >  Z  J5  (Ax.  11).  Q.E.D. 

123.  THEOREM.    If  two  angles  of  a  triangle  are  unequal,  the  sides 
opposite  them  are  unequal  and  the  greater  angle  is  subtended  by  the 
greater  side. 

Given:  A  ABC-, 
To  Prove:    AB  >  AC. 
Proof :    In  Z  ACB,    suppose 
Z  BCR  constructed  =  Z  B. 
Then,  CR=  BR  (?)  (120). 
Also  AR  +  CR  >  AC  (?). 
/.  AR  +  BR  >  ^C  (Ax.  6).    That  is,  AB  >  ^O.  Q.E.D. 

124.  THEOREM.   The  hypotenuse  is  the  longest  side  of  a  right  tri- 
angle.    (See  123.) 


BOOK  I  45 


QUADRILATERALS 

A  quadrilateral  is  a  portion  of  a  plane  bounded 
by  four  straight  lines.  These  four  lines  are  called  the 
sides.  The  vertices  of  a  quadrilateral  are  the  four  points 
at  which  the  sides  intersect.  The  angles  of  a  quadrilateral 
are  the  four  angles  at  the  four  vertices.  The  diagonal  of 
a  rectilinear  figure  is  a  line  joining  two  vertices,  not  in 
the  same  side. 

126.  A  trapezium  is  a  quadrilateral  having  no  two  sides 
parallel. 

A  trapezoid  is  a  quadrilateral  having  two  and  only  two 
sides  parallel. 

A  parallelogram  is  a  quadrilateral  having  its  opposite 
sides  parallel  (O). 


TRAPEZOID  PARALLELOGRAM  SQUARE  RECTANGLE 

RHOMBOID 

127.  A    rectangle  is   a   parallelogram    whose   angles    are 
right  angles. 

A  rhomboid  is  a  parallelogram  whose  angles  are  not  right 
angles. 

128.  A  square  is  an  equilateral  rectangle. 
A  rhombus  is  an  equilateral  rhomboid. 

129.  The  side  upon  which  a  figure    appears   to  stand  is 
called  its  base.     A  trapezoid  and  all  kinds  of  parallelograms 
are  said  to  have  two  bases,  —  the  actual  base  and  the  side 
parallel  to  it.    The  non-parallel  sides  of  a  trapezoid  are  some- 
times called  the  legs.     An  isosceles  trapezoid  is  a  trapezoid 


46 


PLANE   GEOMETRY 


whose  legs  are  equal.  The  median  of  a  trapezoid  is  the 
line  connecting  the  midpoints  of  the  legs.  The  altitude 
of  a  trapezoid  and  of  all  kinds  of  parallelograms  is  the 
perpendicular  distance  between  the  bases. 

130.    THEOREM.    The  opposite  sides  of  a  parallelogram  are  equal. 
Given  :   O  LMOP. 

PO   and 


To  Prove  :    LM 
LP  =  MO. 

Proof  :    Draw  diagonal  PH. 
In  A  LMP  and  OMP,  PM  = 


..--' 


/.  A  LMP  =  A  OMP  (?)    (54). 

/.  LM  =  PO  and  LP  =MO  (?)    (27).  Q.E.D. 

131.  COR.    Parallel  lines  included  between  parallel  lines  are  equal. 
(See  130.) 

132.  COR.    The  diagonal  of  a  parallelogram  divides  it  into  two 
equal  triangles. 

133.  COR.  The  opposite  angles  of  a  parallelogram  are  equal.  (See  27.) 

134.  THEOREM.   If  the  opposite  sides  of  a  quadrilateral  are  equal, 
the  figure  is  a  parallelogram.      [Converse  of  130.] 

Given:  Quadrilateral  ABCD; 
AB  =  DC  ;   AD  =  BC. 

To  Prove :   ABCD  is  a  O. 


Proof  :  Draw  diagonal  BD. 
In  A  ABD  and  CBD,  BD  = 
BD  (?)  ;  AB  =  DC  (?),  and 
AD  =  BC  (?).  .'.  A  ABD  =  A  CBD  (?)  (58). 

Hence  Z  a  =  Z  i  (?).     Therefore  AB  is  ||  to  DC  (?)  (101). 

Also,  Z  y  =  Z.x    (?).     Therefore  AD  is  ||  to  BC  (?). 

Hence  ABCD  is  a  parallelogram  (Def.   126).  Q.E.D. 


BOOK   I  47 

135.   THEOREM    If  two  sides  of  a  quadrilateral  are  equal  and  parallel, 
the  figure  is  a  parallelogram. 

Given:    Quadrilateral  ABCD  ;  AB  =  CD  and  AB  \\  to  CD. 
To  Prove:    A  BCD  is  a  O. 

Proof:    Draw  diagonal   BD.       In   A  ABD  and   C'fi/),  BD  = 
#D(?);  ^47*  =  <?/>  (?);  and  Za  =  Z*  (?)  (97). 
.-.A  ABD  =  A  CtfD  (?)  (52). 
Hence  Zy=Z.r    (?).     .-.   AD   is  ||  to  BC  (?)  (101). 

is  a  parallelogram  (?)   (126).  Q.E.D. 


136.  COR.  Any  pair  of  adjoining  angles  of  a  parallelogram  are  sup- 
plementary.   (See  100.) 

137.  THEOREM.  The  diagonals  of  a  parallelogram  bisect  each  other. 

Given:    O  EFGH  ;    diago-  F- 

nals  EG  and  FH  intersecting 
at  X. 

To  Prove  :    FX  =  X/T   and 
GX  =  ^E. 


Proof  :  In  A  FXG  and  .EXtf,      E 

FG  =  ^H  (?)   (130)  ;  Z  a  =  Z  o  and  Z  c  =  Z  r  (?)   (97). 
.'.A  F^G  =  A  #Xff  (?)  (54). 
.-.  FX  =  XH  and  GX  =  XE  (?)  (27).  Q.E.D. 

138.  THEOREM.  If  the  diagonals  of  a  quadrilateral  bisect  each  other, 
the  figure  is  a  parallelogram. 

Given:  (?).  To  Prove:  (?).  Proof:  In  A  FXG  and 
EXH  show  three  parts  of  one  =  etc.  Hence  certain  A  are 
=  (?).  Then  two  lines  are  II  (?).  Also  =  (?).  Now  use  135. 


Ex.  1.   In  the  figure  of  137,  if  Z  a  =  20°  and  Z  c  =  30°,  find  the  four 
angles  at  X. 

Ex.  2.   If  one  angle  of  a  parallelogram  is  65°,  find  the  other  three.     If 
one  is  90°,  find  the  others. 

Ex.  3.   State  and  prove  the  converse  of  136. 


48  PLANE   GEOMETRY 

139.  THEOREM.  Two  parallelograms  are  equal  if  two  sides  and  the 
included  angle  of  one  are  equal  respectively  to  two  sides  and  the  in- 
cluded angle  of  the  other. 

B _c  M N 


A —  — D  L —  O 

Given  :    HJ  AC  and  LN ;  AB  =  LM  ;   AD  =  LO  ;  Z  A  =  Z  L. 
To  Prove :   The  UJ  are  = . 

Proof:    Superpose    O  ABCD  upon  O  LMNO,  so  the    equal 
angles  A  and  L  coincide,  4D  falling  along  LO  and  ^4B  along  LM. 
Point  D  will  coincide  with  point  O  [^4D  =   LO  (Hyp.)]. 
Point  B  will  coincide  with  point  Jf  [AB  =  £j»f  (Hyp.)]. 
BC   and   MN   are  both  ||  to   LO  (?)   (126). 
/.  BC  falls  along  JOT  (?)  (92). 
CD    and    2VO    are    both  ||  to    LM   (?). 
/.CD  falls  along  .zvo  (?). 
Hence  C  will  fall  exactly  upon  N  (38). 
/.  the   figures  coincide,  and  are  equal  (?)  (28).       Q.E.D. 

140.  THEOREM.  Two  rectangles  are  equal  if  the  base  and  altitude  of 
one  are  equal  respectively  to  the  base  and  altitude  of  the  other.  (See  139.) 

141.  THEOREM.     The  diagonals  of  a  rhombus  (or  of  a  square)  are 
perpendicular  to  each  other,  bisect  each  other,  and  bisect  the  angles  of 
the  rhombus  (or  of  the  square). 

Given :    Rhombus  ABCD  ;   di- 
agonals AC  and  BD. 

To  Prove:   AC  _L   to  BD; 
and  BD  bisect  each  other ; 
they  bisect  A  DAB,  ABC,  etc.      A  B 

Proof  :   Point  A  is  equally  distant  from  B  and  D  (?)   (128). 
Point  C  is  equally  distant  from  B  and  D  (?). 

.'.^Cis±toBD  (?)   (70).  Q.E.D. 


BOOK    T 


49 


Also  AC  and  BD  bisect  each  other  (?)  (137).  Q.E.D. 

The  A  at  A  are  =  (?)  (66,  II).     Etc.  Q.E.D. 

The  proof  if  the  figure  is  a  square  is  exactly  the  same. 

142.   THEOREM.    The  line  joining  the  midpoints  of  two  sides  of  a 
triangle  is  parallel  to  the  third  side  and  equal  to  half  of  it. 

Given:  A  ABC;    M,  the  B 

midpoint  of  AB ;    P,  the 
midpoint  of  BC;  line  MP. 


To  Prove  :  MP 

and  MP  =     AC. 


to  AC 


M 


Proof:     Suppose  AR  is 
drawn  through  A,  II  to  BC         *~ 

and  meeting  MP  produced  at  B.  In  A  ABM  and  BPM,  AM 
=  BM  (Hyp.);  Zx  =  Ze  (?)  and  Z  o  =  Z  B  (?)  (97). 
.'.  A  ABM  =  A  BPM  (?)  (54). 

Hence,  AR  =  BP  (?).     But  BP  =  PC  (?).    .-.  AR  =  PC  (?). 

.-.  ACPR  is  a  O  (?)  (135). 

Hence  RP  or  MP  is  II  to  AC  (?).  Q.E.D. 

Also,  RP  =  AC  (?)  (130).    But  MP  =  RM  (?)  (27). 

.-.  MF  =  J  RP=\  ^C(Ax.  6).  Q.E.D. 

143.   THEOREM.     The   line  bisecting  one  side  of  a  triangle  and 
parallel  to  a  second  side,  bisects  also  the  third  side. 

Given:  A  ABC-,  MP  bisecting 
AB  and  II  to  AC. 

To  Prove  :  MP  bisects  BC  also. 


Proof :  Suppose  MX  is  drawn 
from  M,  the  midpoint  of  AB  to 
X,  the  midpoint  of  BC. 

MX  is  II  to  AC  (142)  ;    but  MP    A 
is  II  to  AC  (Hyp.). 

.-.  MX  and  MP  coincide  (?). 

That  is,  MP  bisects  BC. 

BOBBINS'  PLANE  GEOM.  —  4 


Q.E.D. 


r>0  PLANE    GEOMETRY 

144.  THEOREM.  The  line  bisecting  one  leg  of  a  trapezoid  and 
parallel  to  the  base  bisects  the  other  leg,  is  the  median,  and  is  equal 
to  half  the  sum  of  the  bases. 

Given:  Trapezoid  ABCD\  M,  the  midpoint  of  AB;  MP  II  to 
AD,  meeting  CD  at  P. 

To  Prove:    I.    P  is   the 
midpoint  of  CD. 

II.  MP  is  the  median. 
III.    3fP= 


Proof:  I.   Draw  diagonal 
BD,  meeting  MP  at  R. 

MP  is  II  to  BC  (94).    In  A  ABD,  MR  bisects  BD  (143). 
In  ABDC,  RP  bisects  CD  (?)  (143). 
That  is,  P  is  the  midpoint  of  CD. 
II.    MP  is  the  median  (Def.  129). 
III.    MR  =  1  AD  (142)  and  RP  =  \  BC  (?). 
/.  MP  =  I  {AD  +  J3C)  (Ax.  2).  Q.E.D. 

145.  THEOREM.     The  angles  adjoining  each  base  of  an  isosceles 
trapezoid  are  equal. 

Given  :    Trapezoid   AC  ;   AB  =  CD. 

To  Prove  :    Z  A  =  Z  D   and 

Z  ABC  =  Z  C. 

Proof  :  Suppose  BX  is  drawn 
through  B  and  II  to  CD.  BX  = 
CD  (130);  AB  =  CD  (Hyp.).  A~  ""x""  ~b 

.-.  AB  =  BX  (?),  and  Z  A  =  Z  a  (?),  and  Z  a  =  Z  D  (?)  (98). 

/.Z  A  =  Z  D  (Ax.  1). 

Again,  Z  O  is  supp.  of  Z  D  (?).   Etc.  Q.E.D. 

146.  THEOREM.     If  the  angles  at  the  base  of  a  trapezoid  are  equal, 
the  trapezoid  is  isosceles. 

Given:   (?).     To  Prove:  (?). 

Proof:     Suppose  BX     is  drawn  II  to  CD.     Z  a  =  Z  D  (?); 
Z  A  =  Z  D    (?).      /.  Z  A  =  Z  a  (?).      .'.  AB  =  .BJT  (?).    Etc. 


BOOK   I  51 

NOTE.  The  verb  "to  intersect"  means 
merely  "  to  cut."  In  geometry,  the  verb  ""to 
intercept "  means  "  to  include  between"  Thus 
the  statement "  A  B  and  CD  intercept  X  Y  on 
the  line  EF"  really  means,  " AB  and  CD  A' 
intersect  EF  and  include  XY,  a  part  of  EF, 
between  them." 

147.   THEOREM.     Parallels  intercepting  equal  parts  on  one  trans- 
versal intercept  equal  parts  on  any  transversal. 

Given :   Us  AB,  CD,  EF,  GH,  U  intercepting  equal  parts  AC, 

CE,  EG,  GI,  on  the  transversal  A/ \B 

AI,  and  cutting  transversal  BJ.  I  \ 

To  Prove  :    BD  =  DF  =  FH  c/ \D 


Proof:  The  figure  ABFE  is 
atrapezoid  (?).  CD  bisects  AE 
and  is  II  to  EF  (Hyp.). 

.'.  D  is  midpoint  of  BF  (?).    _ 
That  is,  BD  =  DF. 

Similarly,  CDHG  is  a  trapezoid  and   EF  bisects  DH  (?). 
That  is,  DF=  FH. 

Similarly,  FH  =  HJ.    .  • .  BD  =  DF  =  FH  =  HJ  (Ax.  1).  Q.E.D. 

148.   THEOREM.     The  midpoint  of  the  hypotenuse  of  a  right  triangle 
is  equally  distant  from  the  three  vertices. 

Given  :  Rt.  A  ABC;  M,  the  midpoint  of 
hypotenuse  AB. 

To  Prove :  AM  =  CM  =  BM. 

Proof:  Suppose  MX  is  drawn  II  to  BC, 
meeting  AC  at  X.  X  is  midpoint  of 
AC  (?)  (143).  MX  is  ±  to  AC  (95). 

That  is,  MX  is  J_  to  AC  at  its  midpoint, 
and  AM  =  MC  (?)  (67). 

But    AM=BM    (Hyp.). 

.'.AM=CM  =  BM   (Ax.  1).  Q.E.D.     C 


52  PLANE   GEOMETRY 

149.  THEOREM.     The  median  of  a  trapezoid  is  parallel  to  the  bases 
and  equal  to  half  their  sum. 

[This  is  another  form  of  stating  the  theorem  of  144.] 

150.  THEOREM.   The  perpendiculars  from  the  vertices  of  a  triangle 
to  the  opposite  sides  meet  in  a  point. 


Given  :    A  ABC,  AX J_  to  BC,  BY  J_  to  AC,  and  CZ  _L  to  AB. 
To  Prove :     These  three  Js  meet  in  a  point. 

Proof  :  Through  A  suppose  RS  drawn  II  to  BC  ;  through  B, 
TS  II  to  AC  ;  through  C,  RT  II  to  AB,  forming  A  RST. 

The  figure  ABCR  is  a  O  (Const.)  and  ABTC  is  a  O  (?). 
.'.  RC  =  AB  and  CT  =  AB  (?)  (130).  .*.  RC  =  CT  (Ax.  1). 

Now  cz  is  JL  to  RT  (?)  (95). 

That  is,  CZ  is  _L  to  RT  at  its  midpoint. 

Similarly  AX  is  _L  to  RS  at  its  midpoint. 

And  BY  is  _L  to  TS  at  its  midpoint. 

Therefore  AX,  BY,  CZ  meet  at  a  point  (?)  (85).      Q.E.D. 


Ex.  1.  Draw  the  three  altitudes  of  an  obtuse  triangle  and  prolong 
them  until  they  meet. 

Ex.  2.  Prove  that  each  of  the  three  outer  triangles  in  the  figure  of 
150  is  equal  to  &ABC. 

Ex.  3.  Prove  that  any  altitude  of  A  RST  is  double  the  parallel  alti- 
tude of  A  ,1  BC.  [Use  H3  and  130.] 


BOOK    I  53 

151.  THEOREM    The  point  at  which  two  medians  of  a  triangle  inter- 
sect is  two  thirds  the  distance  from  either  vertex  of  the  triangle  to  the 
midpoint  of  the  opposite  side. 

Given:   A  ABC,  BD  and   CE  two  medians  intersecting  at  O. 
To  Prove  :    HO  =  5  BD  and  CO  =  |  CE. 

<J  O 

Proof :   Suppose  //  is  the  midpoint  of  BO  and  I  is  the  mid- 
point of  CO.     Draw  ED,  D7, 
///,  HE. 

In  A  ABC,  DE  is  II  to  BC 
and  =  i  BC  (?)  (142). 

In   A  OBC,  HI  is  II  to  BC 
and  =  l  BC  (?). 

.-.  ED  =  HI  (Ax.  1),  and 
ED  is  II  to  ///  (?)  (94).  

.-.  EDIH  is  a  O  (?)  (135).    B  c 

.-.  HO  =  oz>  and  10  =OE  (?)  (137). 

.'.  BH=  HO  =  OD  and   C/=  7O=  OJ?  (Ax.  1). 

That  is,  BO  =  2  •  OD  =  f  «Z>,  and  CO=2-EO=%CE.  Q.E.D. 

152.  THEOREM.  The  three  medians  of  a  triangle  meet  in  a  point 
which  is  two  thirds  the  distance  from  any  vertex  to  the  midpoint  of  the 
opposite  side. 

Proof  :  Suppose  AX  is  the  third  median  of  A  ABC  and 
meets  BD  at  o'. 

Then  £Of  =  f  J3D  and  B0  =  f  BD  (?)  (151). 

.-.  BO'  =  BO  (?).  That  is  O'  coincides  with  O  and  the 
three  medians  meet  at  o  which  is  J  the  distance,  etc.  Q.E.D. 

Ex.  1.    In  the  figure  of  151,  prove  EH  =  $  AO  =  D7,by  142. 

Ex.  2.  In  the  figure  of  151,  if  BC>  A  C,  prove  the  angle  EEC  obtuse. 
[Use  87.] 

Ex.  3.  If  one  angle  of  a  rhombus  is  30°,  find  all  the  angles  of  the 
four  triangles  formed  by  drawing  the  diagonals. 

Ex.  4.  Show  that  any  trapezoid  can  be  divided  into  a  parallelogram 
and  a  triangle  by  drawing  one  line. 

Ex.  5.  Prove  that  every  right  triangle  can  be  divided  into  two  isos- 
celes triangles  by  drawing  one  line.  [Use  148.] 


54  PLANE   GEOMETRY 


POLYGONS 

153.  A  polygon    is  a  portion    of    a    plane   bounded    by 
straight  lines.     The  lines  are  called  the  sides.     The  points 
of  intersection  of  the  sides   are   the  vertices.     The   angles 
of  a  polygon  are  the  angles  at  the  vertices. 

154.  The  number  of  sides  of  a  polygon  is  the  same  as  the 
number  of  its  vertices  or  the  number  of  its  angles.      An 
exterior  angle  of  a  polygon  is  an  angle  without  the  polygon, 
between  one  side  of  the  polygon  and  another  side  prolonged. 

155.  An  equilateral   polygon  has   all  of   its   sides   equal 
to   one   another.     An    equiangular   polygon   has    all   of  its 
angles  equal  to  one  another. 

156.  A  convex  polygon  is  a  polygon  no  side  of  which  if 
produced  will  enter  the   surface  bounded  by  the  sides  of 
the  polygon.     A  concave  polygon  is  a  polygon  two  sides  of 
which   if  produced  will  enter  the  polygon. 


EQUILATERAL  EQUIANGULAR  CONCAVE,  OR 

CONVEX  POLYGONS  RE-ENTRANT 

NOTE.  A  polygon  may  be  equilateral  and  not  be  equiangular  ;  or 
it  may  be  equiangular  and  not  be  equilateral.  The  word  "  polygon  " 
is  usually  employed  to  signify  convex  figures. 

157.  Two  polygons  are  mutually  equiangular  if  for  every 
angle  of  the  one  there  is  an  equal  angle  in  the  other  and 
similarly  placed,  Two  polygons  are  mutually  equilateral, 
if  for  every  side  of  the  one  there  is  an  equal  side  in  the 
other,  and  similarly  placed. 


BOOK  I  55 

158.  Homologous    angles    in    two    mutually    equiangular 
polygons  are  the  pairs  of  equal  angles.     Homologous  sides 
in  two  polygons  are  the  sides  between  two  pairs  of  homolo- 
gous angles. 

159.  Two  polygons  are  equal  if  they  are  mutually  equi- 
angular and  their  homologous  sides  are  equal  ;    or  if  they 
are  composed  of  triangles,  equal  each  to  each  and  similarly 
placed.     (Because  in  either  case  the  polygons  can  be  made 
to  coincide.) 

160.  Two   polygons   may  be   mutually  equiangular  without  being 
mutually  equilateral ;  also,  they  may  be  mutually  equilateral  without 
being  mutually  equiangular  —  except  in  the  case  of  triangles. 


The  first  two  figures  are  mutually  equilateral  but  not  mutually  equi- 
angular. The  last  two  figures  are  mutually  equiangular  but  not  mutu- 
ally equilateral. 

161.   A  3-sided  polygon  is  a  triangle. 

A  4-sided  polygon  is  a  quadrilateral. 
A  5-sided  polygon  is  a  pentagon. 
A  6-sided  polygon  is  a  hexagon. 
A  7-sided  polygon  is  a  heptagon. 
An  8-sided  polygon  is  an  octagon. 
A  10-sided  polygon  is  a  decagon. 
A  12-sided  polygon  is  a  dodecagon. 
A  15-sided  polygon  is  a  pentedecagon. 
An  resided  polygon  is  called  an  w-gon. 


Ex.  Draw  a  pentagon  and  all  the  possible  diagonals  from  one  vertex. 
How  many  triangles  are  formed?  Draw  a  decagon  and  the  diagonals 
from  one  vertex.  How  many  triangles  are  thus  formed  ?  Construct  a  20- 
gon  and  the  diagonals  from  one  vertex.  How  many  triangles  are  formed? 


56  PLANE   GEOMETRY 

162.  THEOREM.   The  sum  of  the  interior  angles  of  an  n-gon  is  equal 
to  (n-2)  times  180°. 

Given  :  A  polygon  having 
n  sides. 

To  Prove  :  The  sum  of  its 
interior  A  =  (n  -  2)  •  180°. 

Proof  :  By  drawing  all  pos- 
sible diagonals  from  any  vertex 
it  is  evident  that  there  will 
be  formed  (n  —  2)  triangles. 
The  sum  of  the  A  of  one  A  =  180°  (?)  (110). 

.-.  the  sum  of  the  A  of  (rc-2)  A  =  (w-2)  180°  (Ax.  3). 

But  the  sum  of  the  A  of  the  triangles  =  the  sum  of  the  A 
of  the  polygon  (Ax.  4). 

.-.  sum  of  A  of  the  polygon  =  (n—  2)  180°  (Ax.  1).    Q.E.D. 

163.  COR.   The  sum  of  the  interior  angles  of  an  n-gon  is  equal  to 
iSo°n  -  360°. 

164.  COR.    Each  angle  of  an  equiangular  n-gon  =  1    *  °  • 

MV 

165.  COR.   The  sum  of  the  angles  of  any  quadrilateral  is  equal  to 
four  right  angles. 

166.  COR.   If  three  angles  of  a  quadrilateral  are  right  angles,  the 
figure  is  a  rectangle. 


Ex.  1.  How  many  degrees  are  there  in  each  angle  of  an  equiangular 
pentagon  ?  of  an  equiangular  pentedecagon  ?  of  a  30-gon  ? 

Ex.  2.  If  two  angles  of  a  quadrilateral  are  right  angles,  what  is  true  of 
the  other  two  ? 

Ex.  3.  How  many  sides  has  that  polygon  the  sum  of  whose  interior 
angles  is  equal  to  20  rt.  <4? 

Ex.  4.  How  many  sides  has  that  equiangular  polygon  each  of  whose 
angles  contains  160°? 

Ex.  5.  Tf  in  the  figure  of  105,  Za  =  65°,  how  many  degrees  are  there 
in  each  of  the  other  angles  of  the  figure  ? 


BOOK  1 


57 


167.  THEOREM.    If  the  sides  of  a  polygon  be  produced,  in  order, 
one  at  each  vertex,  the  sum  of  the  exterior  angles  of  the  polygon  will 
equal  four  right  angles,  that  is,  360°. 

Given  :    A  polygon  with  sides 
prolonged  in  succession  forming 
the  several  exterior  angles  a, 
<?,  d,  etc. 

To  Prove  :    Za+Zb  +  Z.  c  + 
Zd  +  etc.  =  4  it.  ^=360°. 

Proof  :    Suppose  at  any  point 
in  the  plane,  lines  are  drawn 
parallel  to  the  several  sides  of  the  given  polygon,  extending 
in  the  same  direction,  and  forming  angles  A,  .B,  C,  D,  etc. 

Then  Z  A  +  /.  B  +  Z.  C  +  Z  D  +  Z  #       \          / 
+  Z  ,F+ZG  =  4  rt.  A  (?)  (47). 

But  Za  =  ZA,  Z5  =  ZU,  Z  c  =  Z 
Z  d=  Z  A  etc.   (?)  (105). 

.'./.a  +  Z  6  +  Z  e  4-  Z  6?  +  Z  e  +  Z/  /       \ 

+  Z#  =  4  rt.  ^  =  360°  (?)    (Ax.  6).  /  \ 

Q.E.D. 

168.  COR.    Each  exterior  angle  of  an  equiangular  polygon  is  equal 

4  rt.  A  360° 

to*   n      ;    that  is,  =^— 

169.  COR.    The  sum  of  the  exterior  angles  of  a  polygon  is  indepen- 
dent of  the  number  of  its  sides. 


c     .,•-•"" 


F 


Ex.  1.  How  many  degrees  are  there  in  each  exterior  angle  of  an 
equiangular  dodecagon  V  of  an  equiangular  40-gon  ? 

Ex,  2.  If  any  angle  of  an  isosceles  triangle  is  60°,  the  triangle  is 
equiangular. 

Ex.  3.  Prove  the  theorem  of  162  by  drawing  lines  from  any  point 
within  the  triangle  to  all  the  vertices. 

Ex.  4.   State  the  theorems  which  deal  with  concurrent  lines. 

Concurrent  lines  are  lines  that  meet  in  a  common  point. 

Ex.  5.  IIo\v  many  sides  has  that  polygon  the  sum  of  whose  interior 
angles  exceeds  the  sum  of  the  exterior  angles  by  720°? 


58 


PLANE  GEOMETRY 


SYMMETRY 

170.  A  figure  is  symmetrical  with  respect  to  a  line  if,  by 
using  that  line  as  an  axis,  the  part  of  the  figure  on  one  side 
of  the  line  may  be  folded  over,  and  will  exactly  coincide  with 
the  part  on  the  other  side.    This  line  is  an  axis  of  symmetry. 

171.  A  figure  is  symmetrical  with  respect  to  a  point  if  this 
point  bisects  every  line  drawn  through  it  and  terminated 
(both  ways)  in  the  boundary  of  the  figure. 

This  point  is  the  center  of  symmetry. 

172.  It  is  evident  that  the  axis  of  symmetry  bisects  at 
right  angles  every  line  joining  two  symmetrical  points ;  and 
that  the  center  of  symmetry  bisects  every  line  joining  any 
pair  of  points  symmetrical  with  respect  to  it. 


Examples  of  symmetry  are  given  in  these  figures. 
First  figure  is  symmetrical  with  respect  to  XX'  as  an  axis.     (Why?) 
Second  figure  is  symmetrical  with  respect  to  0  as  a  center.     (Why?) 
P'  and  /Y  are  symmetrical  with  respect  to  XX'  as  an  axis.     (Why?) 
A  and  A',  B  and  B'  etc.  are  symmetrical  with  respect  to  0  as  a  center. 
(Why?)     XX1  is  J_  to  P'P/  and  bisects  it.     A  0  =  A'O,  BO  =  B'O,  etc. 

173.  In  order  to  prove  that  a  line  is  an  axis  of  symmetry 
it  is  necessary  to  show  that  it  satisfies  the  condition  of  170. 

In  order  to  prove  that  a  point  is  a  center  of  symmetry  it  is 
necessary  to  show  that  it  satisfies  the  condition  of  171. 


BOOK    I 


59 


174.  THEOREM.   If  two  lines  are  symmetrical  with  respect  to  a 
center,  they  are  equal  and  parallel. 

Given  :  AB  and  RS  symmet- 
rical with  respect  to  O,  that  is, 
every  line  through  O,  termi- 
nated in  AB  and  RS  is  bisected 
at  O ;  AS  and  BR>  two  such 
lines. 

To  Prove  :    AB  =  RS  and  AB  II  to  RS. 

Proof :    Draw  AR  and  BS.     AO  =  OS  and  BO  =  OR  (Hyp.). 

.-.   ABSR  is  aO(?)  (138). 

.'.  AB  =  RS  (?)  and  AB  is  II  to  RS  (?).  Q.E.D. 

175.  THEOREM.    If  a  diagonal  of  a  quadrilateral  bisects  two  of  its 
angles,  this  diagonal  is  an  axis  of  symmetry. 

Given :  Quadrilateral 
ABCD-,  AC  a  diagonal  bisect- 
ing Z  BAD  and  Z  BCD. 

To  Prove  :  ABCD  symmet- 
rical with  respect  to  AC. 

Proof :  In  A  ABC  and 
ADC,  AC=  AC  (?). 

Z  BAG  =  Z  DAC  (?)  and 
Z  BCL4  =Z  DCM  (?). 

Hence  A  ABC  =  A  ADC  (?). 

/.  ^4(7  is  an  axis  of  symmetry  (?).  Q.E.D. 

176.  COR.    The  diagonal  of  a  square  or  of  a  rhombus  is  an  axis 
of  symmetry.     (Why  ?) 

177.  COR.    The  diagonal  of  a  rectangle  or  of  a  rhomboid  is  not  an 
axis  of  symmetry.      (Why  not?) 

Ex.  1.  Is  the  altitude  of  an  equilateral  triangle  an  axis  of  symmetry? 

Ex.  2.  Is  the  altitude  of  an  isosceles  triangle  an  axis  of  symmetry? 

Ex.  3.  Are  all  altitudes  of  all  triangles  axes  of  symmetry? 

Ex.  4.  Has  an  isosceles  trapezoid  an  axis  of  symmetry  ? 


60 


PLANE   GEOMETRY 


178.  THEOREM.    If  a  figure  is  symmetrical  with  respect  to  two  per- 
pendicular axes,  it  is  symmetrical  with  respect  to  their  intersection  as  a 
center. 

Given:  Figure  MN  sym- 
metrical with  respect  to  the 
J_  axes  XX1  and  YYr  which 
intersect  at  O. 

To  Prove :    Figure    MN    is    x- 
symmetrical  with  respect  to 
O  as  a  center. 

Proof:  Take  any  point  P 
in  the  boundary.  Draw  PB 
-L  to  FF',  intersecting  YT1  at  A  and  meeting  the  boundary 
at  B.  Draw  BR  J_  to  XX1,  intersecting  XX1  ^  C  and  meet- 
ing the  boundary  at  R.  Draw  AC,  OP,  OR. 

[The  demonstration  is  accomplished    by  proving  ROP  a 
straight  line,  bisected  at  O.] 

PB  is  II  to  XX1  and  BR  is  I!  to  FF;  (?)  (93). 

Hence  ABCO  is  a  O  (?). 

.-.  BC=AO  (?).     But  BC=CR  (?)  (172). 

.'.   AO=  CR  (Ax.  1). 

.'.  ACRO  is  a  O  (?)  (135).     /.  RO  is  =  and  II  to  AC  (?). 

Similarly  CO  is  =  and  II  to  AP ;  hence  ACOP  is  a  O  (?). 

/.  PO  is  =  and  II  to  AC  (?). 

/.  POR  is  a  straight  line  (?)  (92)  and  PO  —  OR  (?). 

But  P  is  any  point,  so  POR  is  any  line  through  O. 

Hence  O  is  a  center  of  symmetry  (171).  Q.E.D. 

LOCUS 

179.  The  locus  of  a  point  is  the  series  of  positions  the  point 
must  occupy  in  order  that  it  may  satisfy  a  given  condition. 
It  is  the  path  of  a  point  whose  positions  are  limited  or  defined 
by  a  given  condition,  or  given  conditions. 

180.  Explanatory.    I.    If  a  point  is  moving  so  that  it  is 
always   one  inch  from  a  given  indefinite  straight  line,  the 


BOOK   I  61 

point  may  occupy  any  position  in  either  of  two  indefinite  lines, 
one  inch  from  the  given  line,  parallel  to  it,  and  one  on  either 
side  of  it.  And  this  point  cannot  occupy  any  position  which 
is  not  in  these  lines-  Hence,  the  locus  of  points  at  a  given 
distance  from  a  given  line  is  a  pair  of  parallels  to  the  given 
line,  one  on  either  side  of  it,  and  at  the  given  distance  from  it. 

II.  If  a  point  is  moving  so  that  it  is  always  equally  distant 
from  two  parallels,  it  must  move  in  a  third  parallel  midway 
between  them.     Hence,  the  locus  of  points  equally  distant 
from  two  parallels  is  a  third  parallel  midway  between  them. 

III.  The  method  of  proving  that  a  certain  line  or  group 
of  lines  is  the  locus  of  points  satisfying  a  given  condition, 
consists  in  proving  that  every  point  in  the  line  fulfills  the 
given   requirement,  and   that  there  is  no   other  point  that 
fulfills  it.     In  the  above  illustrations  it  is  evident  that  every 
point  in  the  lines  which  were  called  the  "locus,"  did  fulfill 
the  conditions  of  the  case.     It  is  also  evident  that  there  is 
no  point  outside   these    "  loci "  which  does   so    fulfill    the 
conditions.     That   is,    these    "loci"    contain  all  the  points 
described  and  no  others. 

IV.  THEOREM.   The  locus  of  points  equally  distant  from  the  ex- 
tremities of  a  line  is  the  perpendicular  bisector  of  the  line. 

Proof:  Every  point  in  the  _L  bisector  of  a  line  is  equally 
distant  from  its  extremities  (67).  And  also,  there  is  no 
point  outside  the  _L  bisector  which  is  equally  distant  from 
the  extremities  of  a  line  (69).  Hence  it  is  the  locus  of 
points  equally  distant  from  the  extremities  of  the  line. 

V.  THEOREM.    The  locus  of  points  equally  distant  from  the  sides 
of  an  angle  is  the  bisector  of  the  angle. 

Proof  :    Like  the  preceding  proof.      [Use  79,  80,  81.] 

VI.  The  locus  of  the  vertices  of  all  the  isosceles  triangles 
that  can  be  constructed  on  a  given  base  is  the  perpendicular 
bisector  of  the  base.      (Same  as  IV.) 


62  PLANE   GEOMETRY 

CONCERNING  ORIGINAL  EXERCISES 

181.  In   the  original  work  which  this  text  contains,  the 
pupil  is  expected  to  state  the  hypothesis  and  conclusion  of 
each  theorem,  and  apply  them  to  an  appropriate  figure.     He 
is   expected   to   state    completely  and  logically    the  proof, 
giving  a  correct  reason  for  every  declarative  statement. 

In  many  of  these  exercises,  suggestions  are  made  and  such 
assistance  is  given  as  experience  has  shown  average  pupils 
require.  This  is  done  in  order  that  the  learner  may  be  en- 
couraged toward  definite  accomplishment,  which  is  one  of 
the  greatest  incentives  to  further  effort. 

To  apply  the  knowledge  acquired  from  the  preceding  pages 
is  now  the  student's  task.  His  fascination  for  this  science 
will  depend  largely  upon  the  success  of  his  efforts  at  proving 
originals.  Therefore,  many  obstacles  will  be  removed  or  modi- 
fied, and  110  trouble  will  be  spared  in  making  the  mastery  of 
this  department  of  geometry  both  agreeable  and  profitable. 

The  student  should  not  draw  a  special  figure  for  a  general 
proposition.  That  is,  if  "triangle  "  is  specified,  he  should  draw 
a  scalene  and  not  an  isosceles  or  a  right  triangle  ;  and  if 
"quadrilateral"  is  mentioned,  he  should  draw  a  trapezium 
and  not  a  parallelogram  or  a  square. 

SUMMARY.     GENERAL  DIRECTIONS  FOR  ATTACKING  EXERCISES 

182.  A  triangle  is  proven  isosceles  by  showing  that  it  contains  two 
equal  sides,  or  two  equal  angles. 

,  183.  A  triangle  is  proven  a  right  triangle  by  showing  that  one  of  its 
angles  is  a  right  angle,  or  two  of  its  angles  are  complementary,  or  one  of 
its  angles  is  equal  to  the  sum  of  the  other  two. 

184.    Right  triangles  are  proven  equal,  by  showing  that  they  have : 

(1)  Hypotenuse  and  acute  angle  of  one  =  etc. 

(2)  Hypotenuse  and  leg  of  one  =  etc. 

(3)  The  legs  of  one  =  etc. 

(4)  Leg  and  adjoining  angle  of  one  =  etc. 

(5)  Leg  and  opposite  angle  of  one  =  etc. 


BOOK  I  63 

185.  Oblique  triangles  are  proven  equal,  by  showing  that  they  have : 

(1)  Two  sides  and  the  included  angle  of  one  =  etc. 

(2)  One  side  and  the  adjoining  angles  of  one  =  etc. 

(3)  Three  sides   of  one  =  etc. 

186.  Angles  are  proven  equal,  by  showing  that  they  are : 

(1)  Equal  to  the  same  or  to  equal  angles. 

(2)  Halves  or  doubles  of  equals. 

(3)  Vertical  angles. 

(4)  Complements  or  supplements  of  equals. 

(5)  Homologous  parts  of  equal  figures. 

(6)  Base  angles  of  an  isosceles  triangle. 

(7)  Corresponding  angles,  alternate-interior  angles,  etc.  of  parallels. 

(8)  Angles  whose  sides  are  respectively  parallel  or  perpendicular. 

(9)  Third  angles  of  triangles  which  have  two  angles  of  one  =  etc. 

187.  Lines  are  proven  equal,  by  showing  that  they  are : 

(1)  Equal  to  the  same  or  to  equal  lines. 

(2)  Halves  or  doubles  of  equals. 

(3)  Distances  to  the  ends  of  a  line  from  any  point  in  its  perpendicu- 

lar bisector. 

(4)  Homologous  parts  of  equal  figures. 

(5)  Sides  of  an  isosceles  triangle. 

(6)  Distances  to  the  sides  of  an  angle  from  any  point  in  its  bisector. 

(7)  Opposite  sides  of  a  parallelogram. 

(8)  The  parts  of  one  diagonal  of  a  parallelogram  made  by  the  other. 

188.  Two  lines  are  proven  perpendicular,  by  showing  that  they  : 

(1)  Make  equal  adjacent  angles  with  each  other. 

(2)  Are  legs  of  a  right  triangle. 

(3)  Have  two  points  in  one,  each  equally  distant  from  the  ends  of  the 

other. 

189.  Two  lines  are  proven  parallel,  by  : 

(1)  The  customary  angle-relations  of  parallel  lines. 

(2)  Showing  that  they  are  opposite  sides  of  a  parallelogram. 

(3)  Showing  that  they  are  parallel  or  perpendicular  to  a  third  line. 

190.  Two  lines,   or  two  angles,  are  proven  unequal  by  the  usual 
axioms  and  theorems  pertaining  to  inequalities. 

[See  especially,  Ax.  5*;  Ax.  12;  68;  75;  76,  III;  77;  86  ;  87*;  90 
109*;  122*;  124.] 

*  These  refer  to  angles. 


04  PLANE   GEOMETRY 


ORIGINAL   EXERCISES 

1.  A  line  cutting  the  equal  sides  of  an  isosceles  triangle  and  parallel 
to  the  base  forms  another  isosceles  triangle.     [Use  186  (7),  and  182.] 

2.  The  bisectors  of  the  equal  angles  of  an  isosceles  triangle  form,  with 
the  base,  another  isosceles  triangle.     [Use  186  (2).] 

H 

3.  If  the  exterior  angles  at  the  base  of  a  triangle 
are  equal,  the  triangle  is  isosceles.    [186  (4).] 

4.  If  from  any  point  in  the  base  of  an  isosceles  tri- 
angle a  line  be  drawn  parallel  to  one  of  the  equal 
sides  and  meeting  the  other  side,  an  isosceles  triangle    .< 
will  be  formed. 

To  Prove  :  A  DEC  isosceles. 


5.  If  the  median  of  a  triangle  is  perpendicular 
to  the  base,  the  triangle  is  isosceles.     [Use  187  (3).] 

6.  If  a  line  through  the  vertex  of  a  triangle 
and  parallel  to  the  base,  makes  equal  angles  with 
the  sides,  the  triangle  is  isosceles. 

Given  :  /La  =  Z  x,  etc. 

7.  The  median  of  an  isosceles  triangle  is  per- 
pendicular to  the  base.     [Use  188  (3).] 

8.  The  bisectors  of   two  supplementary-adja- 
cent angles  are  perpendicular  to  each  other. 

Proof:   Z.40£+ZjB<9C  =  180°(?);  %Z.AOB  + 
=9Q°  (Ax.  3).  \  ^AOB  =  ZROB;  etc. 


9.    The    bisectors    of     two    adjoining-  interior 
angles  of  two  parallels  meet  at  right  angles. 

Proof:    /.MAC  +  /.MCA  =  90°  as  in  No.  8. 
Then  use  183. 

10.  If  the  bisector  of  an  exterior  angle  of  a  triangle 
is  parallel  to  the  base,  the  triangle  is  isosceles. 

11.  If  the  sum  of  two  angles  of  a  triangle  is  equal  to 
the  third,  the  triangle  is  a  right  triangle. 

[Use  110.] 


BOOK    I 


65 


12.  If  the  median  of  a  triangle  is  equal  to  half  the  side  to  which  it  is 
drawn,  it  is  a  right  triangle. 

Given  :  MA  =  MB  =  MC.  c 

To  Prove  :  A  ABC  a  rt.  A. 
Proof  :  /.A  =  /.ACM  (?)  ;  Z.E  =  /.BCM  (•?). 
/.  by  adding,  etc.   (Use  Ax.  2  and  183.)  A*" 

13.  If  from  any  point  in  the  bisector  of  an  angle 
a  line  be  drawn  parallel  to  either  side  of  the  angle,  an 
isosceles  triangle  will  be  formed.    [182.] 

14.  If   the  bisector  of   the  vertex-angle  of  a  tri- 
angle is  perpendicular  to  the  base,  the  triangle  is 
isosceles. 

Proof:   The  rt.  &are=  [184  (4)].     Then  use  187 
(4);  etc. 

15.  Every  isosceles  right  triangle  can  be  divided 
by  one  line  into  two  isosceles  right  triangles. 


M 


16.  The  diagonals  of  a  rhombus  divide  the  figure  into  four  equal 
right  triangles.     [141.] 

17.  If  a  line  be  drawn  perpendicular  to  the  bisec- 
tor of  an  angle  terminating  in  the  sides,  the  right 
triangles  formed  will  be  equal. 

18.  If  from  each  point  at  which  a  transversal  inter- 
sects two  parallels  a  perpendicular  to  the  other  paral- 
lel be  drawn,  two  equal  right  triangles  will  be  formed. 

19.  If   two    perpendiculars   be   drawn   to    the 

upper  base  of  a  parallelogram  from  the  extremi-        R__D T     C 

ties  of  the  lower  base,  two  equal  right  triangles 
will  be  formed. 


20.  The  perpendiculars  to  the  equal  sides  of  an 
isosceles  triangle  from  the  opposite  vertices  form 
two  pairs  of  equal  right  triangles. 

21.  If  two  intersecting  lines  have  their  extremities 
in  two  parallels  and  their  point  of  intersection  bisects 
one  of  them,  it  bisects  the  other  also. 

Given:    AO  =  EO-,  etc. 

BOBBINS'  PLANE  GEOM.  —  5 


C 


66 


PLANE  GEOMETRY 


22.  If  two  adjacent  sides  of  a  quadrilateral  are 
equal   and   the    diagonal   bisects    their    included 
angle,  the  other  two  sides  are  equal. 

23.  If  a  diagonal  of  a  quadrilateral  bisects  two 
of   its  angles,  the  quadrilateral  has  two  pairs  of 
equal  sides. 

24.  The  altitudes  of  an  isosceles  triangle  upon  the  legs  are  equal. 

25.  If   a  triangle    has  two   equal   altitudes,   it  is 
isosceles. 

26.  The  diagonals  of  a  rectangle  are  equal. 

27.  The  medians  drawn  from  the  ends  of  the  base  of 
an  isosceles  triangle  are  equal. 

28.  The   three  lines  joining  the  midpoints  of  the 
sides  of  a  triangle   divide  the  triangle  into  four 
equal  triangles.     [Use  142  and  132.] 

29.  The     bisector    of  the   vertex-angle  of    an 
isosceles  triangle  bisects  the  base  at  right  angles. 
[Use  185(1);  27;  50.] 


30.  The  bisectors  of  the  equal  angles  of  an  isos- 
celes triangle  (terminating  in  the  equal  sides)  are  equal.     [Use  185  (2).] 

31.  The  median  to  the  base  of  an  isosceles  triangle  bisects  the  vertex- 
angle.     [Use  185  (3).] 

32.  The  diagonals  of  an  isosceles  trapezoid  are 
equal.     [Use  145.] 

33.  The  diagonals  of  an  isosceles  trapezoid  divide 
the  figure  into  four  triangles,  of  which  one  pair  is 
isosceles  and  the  other  pair  is  equal. 


34.  What    is  the   complement   of  an  angle  containing   35°?    80°? 

75°  25'?  8°  18'? 

35.  What  is  the  supplement  of  50°?  100°?   148°?    113°  48'  ? 

36.  In  a  right  triangle  ABC,  if  Z.  A  is  47°,  find  Z  B. 

37.  In  an  isosceles  triangle  /.  A  =  Z.  B  =  80° ;  find  Z  C. 

38.  In  A  ABC,  if  Z  A  =  25°,  Z  B  =  88°,   find  Z  C.    Find  the  exterior 
angle  at  A . 


BOOK   I  67 

39.  In  A  ABC,  \i^.A-  40°,  Z.  B  =  70°  40' ;  find  £  C  and  the  exterior 
angle  at  B. 

40.  The  vertex-angle  of  an  isosceles  triangle  is  44°.    Find  each  base 
angle. 

41.  How  many  degrees  are  there  in  the  sum  of  the  angles  of  a  penta- 
gon ?   of  a  decagon  ?  of  a  9-gon  ? 

42.  How  many  degrees  are  there  in  each  angle  of  an  equiangular  hexa- 
gon ?  of  an  equiangular  octagon  ? 

43.  How  many  degrees  are  there  in  each  exterior  angle  of  an  equian- 
gular pentagon  V    hexagon  ?    dodecagon  V    16-gon  ? 

44.  If  one  acute  angle  of  a  right  triangle  is  double  the  other,  how 
many  degrees  are  there  in  each  ?    [Denote  the  less  Z  by  x.~] 

45.  If  the  acute  angles  of  a  right  triangle  are  equal,  how   many 
degrees  are  there  in  each  ? 

46.  If  one  acute  angle  of  a  right  triangle  is  five  times  the  other,  how 
many  degrees  are  there  in  each  ? 

47.  If  a  base  angle  of  an  isosceles  triangle  is  60°,  find  the  vertex-angle. 
What  kind  of  triangle  is  this  ? 

48.  If  the  vertex-angle  of  an  isosceles  triangle  is  60°,  find  each  base 
angle.     What  kind  of  triangle  is  this  ? 

49.  If  the  vertex-angle  of  an  isosceles  triangle  equals  twice  the  sum 
of  the  two  base  angles,  how  many  degrees  are  there  in  each  angle? 

50.  If  the  vertex-angle  of  an  isosceles  triangle  equals  four  times  the 
sum  of  the  base  angles,  find  each  angle. 

51.  If  the  vertex-angle  of  an  isosceles  triangle  is  half  each  of  the  base 
angles,  find  each  angle. 

52.  If  one  angle  of  a  parallelogram  is  54°,  how  many  degrees  are 
there  in  each  of  the  remaining  angles? 

53.  If  a  transversal  cuts  two  parallels  making  one  pair  of  alternate- 
interior  angles  each  40°,  how  many  degrees  are  there  in  each  of  the  other 
six  angles  formed  ? 

54.  Find  the  three  angles  formed  by  the  bisectors  of  the  angles  of  a 
triangle  whose  angles  are  44°,  62°,  and  74°. 

55.  If  Z  A  of  A  ABC  is  33°  and  the  exterior  angle  at  C  is  110°, 
find  Z  B. 


68  PLANE   GEOMETRY 

56.  If  two  angles  of  a  triangle  are  80°  and  55°,  how  many  degrees  are 
there  in  the  angle  formed  by  their  bisectors? 

57.  The  vertex-angle  of  an  isosceles  triangle  is  one  third  of  either  ex- 
terior angle  at  the  extremities  of  the  base.     Find  each  angle  of  the 
triangle. 

58.  If  two  angles  of  a  triangle  are  30°  and  40°, 
how  many  degrees  are  there  in  the  angle  formed 
by  the  bisector  of  the  third  angle  and  the  altitude 
from  the  same  vertex  ?     Solution  :  Z  x  =  Z  A  BS  — 
£ABD=\£  ABC  -  comp.  of  Z  A. 

59.  Theorem.    The  angle  between  the  altitude  of  a  triangle  and  the 
bisector  of  the  angle  at  the  same  vertex  equals  half  the  difference  of  the 
other  angles  of  the  triangle. 

Proof:    Z  x  =  Z  ABS  -  Z  ABD  =  \   (180°  - 
/.A  -  Z  C)  -  (90°  -  Z  A)  =  etc. 

60.  Theorem.    The  exterior  angle  at  the  base 
of  an  isosceles  triangle  equals  half  the  vertex- 
angle  plus  90°.    Proof  :  Zz  =  Za  +  Zr  =  etc. 

61.  If  in  A  ABC,  Z  BAG  =  80°,  Z  ABC  =  30°, 
find  the  angle  formed  by  the  bisectors  of  the  exte- 
rior angles  at  A   and  B.      Solution  :  Z  x  =  180°  - 
/.BAD-/.  ABD]    Z  BAD  =  \(\m°-  Z.A};    etc. 


62.  The  angle  formed  by  the  bisectors  of  two  ex- 
terior angles  of  a  triangle  equals  half  the  sum  of 
the  interior  angles  at  the  same  vertices. 

63.  If  one  acute  angle  of  a  right  triangle  is  double 
the  other,  the  hypotenuse  is  double  the  shorter  leg. 

[Denote  the   less  Z  by  x.     Find  the  other.     Draw  the  median  from 
vertex  of  rt.  Z.     Prove  one  A  formed  equilateral.] 

64.  If  one  angle  of  a  triangle  is  double  another, 
the  line  from  the  third  vertex,  making  with  the 
longer  adjacent  side   an    angle  equal  to  the  less 

given  angle,  divides  the  triangle  into  two  isosceles  triangles. 


65.    The  bisector  .of  an  angle  bisects,  if  pro- 
duced, the  vertical  angle  also. 
Given:  Z  DON  =  Z  BON. 


BOOK    I 


69 


66.  A  line  perpendicular  to  the  bisector  of   an  angle  at  the  vertex 
bisects  its  supplementary-adjacent  angle.  _ 

Given  :  Z  a=  /.  b  and  OF  ±  to  ON. 
(Use  48.) 

67.  If  the  bisectors  of  two  adjacent  angles 

are  perpendicular,  the  angles  are  supplementary.     A" 
Given:  Za=Z6;  Zar  =  Zz; 


68.  The  bisectors  of  any  two  adjoining  angles  of  a 
parallelogram  meet  at  right  angles.      [Use  136 ;  183.] 

69.  If  from  any  point  in  the  base  of  an  isosceles 
triangle  perpendiculars  to  the  equal  sides  be  drawn, 
they  will  make  equal  angles  with  the  base. 

[Use  114;  48.] 

70.  If  a  line  be  drawn  through  the  vertex  of  an 
angle  and  perpendicular  to  the  bisector  of  the  angle, 
it  will  make  equal  angles  with  the  sides. 

To  Prove :.  ^  r  =  z  s. 

71.  The  bisector  of  the  exterior  angle  at  the  ver- 
tex of  an  isosceles  triangle  is  parallel  to  the  base. 

Proof:   Z  DCB  =  2  /.A   (?)  and  =2  Z.  DCR  (?). 
Etc. 

72.  The  line  through  the  vertex  of  an  isosceles  tri- 
angle, parallel  to  the  base,  bisects  the  exterior  angle. 

73.  Parallel  lines  are  everywhere  equally  distant. 
Given :  ||.  A C  and  BD;  AB  and  CD  &  to  A C. 

To  Prove :  AB  =  CD.     (Use  93;  130.) 

74.  If  two  lines  in  a  plane  are  everywhere  equally 
distant,  they  are  parallel.     [Use  93 ;  135.] 

75.  If  the  diagonals  of  a  parallelogram  are  equal, 
the  figure  is  a  rectangle. 

[Use  A  ABC  and  DBC;  185  (3);  50.]  ' 


76.  The  perpendiculars  upon  a  diagonal  of  a 
parallelogram  from  the  opposite  vertices  are 
equal.  [184  (1).]  A' 


TO  PLANE   GEOMETRY* 

77.  The  perpendiculars  to  the  legs  of  an  isosceles  triangle  from  the 
midpoint  of  the  base  are  equal. 

78.  State  and  prove  the  converse  of  No.  77. 

79.  If  AB  =  LM  and  AL  =  EM,  Z  B  =  Z  L 
audZBAO  =  ^OML2MdBO=OL. 

80.  Any  line  terminated  in  a  pair  of  oppo-  c 

site    sides    of    a    parallelogram    and    passing  " 

through  the  midpoint  of  a  diagonal  is  bisected 
by  this  point.    To  Prove :  RO  =  OS. 

81.  The  midpoint  of  a  diagonal  of  a  paral- 
lelogram is  a  center  of  symmetry. 

82.  If  the  base  angles  of  a  triangle  be  bisected 
and  through  the  intersection  of  the  bisectors  a  line 
be  drawn  parallel  to  the  base  and  terminating  in  the 
sides,  this  line  will  be  equal  to  the  sum  of  the  parts 

of  the  sides  it  meets,  between  it  and  the  base.  A  ""^c 

83.  In  two  equal  triangles,  homologous  medians  are  equal.     Homolo- 
gous altitudes  are  equal.     Homologous  bisectors  are  equal. 

84.  If  two  parallel  lines  are  cut  by  a  transversal,  the  two  exterior 
angles  on  the  same  side  of  the  transversal  are  supplementary. 

85.  If  from  a  point  a  perpendicular  be  drawn  to  each  of  two  parallels 
they  will  be  in  the  same  line.     [Draw  a  third  II  through  the  point.] 


86.  One  side  of  a  triangle  is  less  than  the  sum  of  the  other  two  sides. 

87.  The  sum  of  the  sides  of  any  polygon  ABODE  is  greater  than  the 
sum  of  the  sides  of  triangle  A  CE. 

88.  If  X  is  a  point  in  side  AB  of  A  ABC,  AB  +  BOAX  +  XC. 

89.  In  the  figure  of  No.  88  Z  AXO  Z  B. 

90.  If  lines  be  drawn  from  any  point  within  a  triangle  to  the  ends  of 
the  base,  they  will  include  an  angle  which  is  greater  than  the  vertex 
angle  of  the  triangle.     [Use  109  with  figure  of  75.] 

91.  Any  point  (except  the  vertex)  in  either  leg  of  an  isosceles  triangle 
is  unequally  distant  from  the  ends  of  the  base.  B 

92.  If  two  sides  of  a  triangle  are  unequal  and 
the  median  to  the  third  side  be  drawn,  the  angles 
formed  with  the  base  will  be  unequal.     [Use  87.] 

93.  State  and  prove  the  converse  of  No.  92. 


BOOK    I 


71 


94.  If  the  side  LM,  of  equilateral  triangle  LMN, 
be  produced 'to  />,  and  PN  be  drawn,  Z  PNL  >  £  L  > 
Z  P.     Also  PL>PN>  LN. 

95.  If  from  any  point  within  a  triangle  lines  be 
drawn  to  the  three  vertices: 

(1)  Their  sum  will  be  less  than  the  sum  of  the  sides 
of  the  triangle.     [Use  75  three  times.] 

(2)  Their  sum  will  be  greater  than  half  the 
sum  of  the  sides  of  the  triangle.     [Use  Ax.  12 
three  times.] 

96.  The  sum  of  the  diagonals  of  any  quadri- 
lateral is  less  than  the  sum  of  the  four  sides; 
but  greater  than  half  that  sum. 

97.  The  line  drawn  from  any  point  in  the  base  of  an   isosceles 
triangle  to  the  opposite  vertex  is  less  than  either  leg. 


98.  The  bisectors  of  a  pair  of  corresponding  angles  are  parallel. 
[Use  98 ;  189,  etc.] 

99.  If  two  lines  are  cut  by  a  transversal  and  the  exterior  angles  on 
the  same  side  of  the  transversal  are  supplementary,  the  lines  are  parallel. 

100.  The  bisectors  of  a  pair  of   vertical   angles  are   in    the  same 
straight  line. 

101.  If  one  angle  of  a  parallelogram  is  a  right  angle  the  figure  is  a 
rectangle. 

102.  The  bisectors  of  the  angles  of  a  trapezoid  form  a  quadrilateral 
two  of  whose  angles  are  right  angles.  ^ 

103.  The  bisectors    of    the  four  interior      -—  ^/ 
angles   formed   by  a  transversal  cutting  two 

parallels  form  a  rectangle. 

[Prove  each  Z.  of  LMPQ  a  rt.  ^.] 

104.  The  bisectors  of  the  angles  of  a  par- 
allelogram form  a  rectangle. 

105.  The  bisectors  of  the  angles  of  a  rectangle 
form  a  square.     [In  order  to  prove  EFGH  equilat- 
eral, the  &  AHB  and  CDF  are  proven  equal  and 
isosceles;  similarly  & BG C  and  AED.] 


72 


PLANE   GEOMETRY 


106.  The    lines  joining  a  pair  of   opposite 
vertices  of  a  parallelogram  to  the  midpoints  of 
the  opposite  sides  are  .  }ual  and  parallel.    [Prove 
BCEF  a  O.] 

107.  If  the  four  midpoints  of  the  four  halves 
of  the  diagonals  of  a  parallelogram  be  joined 
in  order,  another  parallelogram  will  be  formed. 

108.  If  the  points  at  which  the  bisectors 
of   the  equal  angles  of   an  isosceles  triangle 
meet  the  opposite  sides,  be  joined  by  a  line,  A 
it  will  be  parallel  to  the  base. 

109.  If  two  angles  of  a   quadrilateral  are   supple- 
mentary, the  other  two  are  supplementary.    [Use  165.] 

110.  If  from  any  point  in  the  base  of  an  isosceles 
triangle  parallels  to  the  equal  sides  be  drawn,  the  sum 
of  the  sides  of  the  parallelogram  formed  will  be  equal 
to  the  sum  of  the  legs  of  the  triangle. 

To  Prove:  XY  +  YC  +  CZ  +  XZ  =  AC  +  EC. 

111.  If  one  of  the  legs  of  an  isosceles  triangle  be 
produced  through  the  vertex  its  own  length,  and  the 
extremity  be  joined  to  the  nearer  end  of  the  base, 
this  line  will  be  perpendicular  to  the  base.    [Use  183.] 

112.  If  the  middle  point  of  one  side  of  a  triangle 
is  equally  distant  from  the  three  vertices,  the  triangle 
is  a  right  triangle. 

[Proof  and  figure  same  as  for  No.  111.] 

113.  If  through  the  vertex  of  the  right 
angle  of    a  right  triangle  a  line  be  drawn 
parallel  to  the  hypotenuse,  the  legs  of  the 
right  triangle  wiU  bisect  the  angles  formed 
by  this  parallel   and  the  median  drawn  to 
the  hypotenuse.     [Use  148 ;  97 ;  etc.]  -A 

114.  Any  two   vertices    of    a    triangle    are 
equally  distant  from  the  median  from  the  third 
vertex. 

115.  If  from  any  point  within  an  angle  per- 
pendiculars  to  the   sides  be  drawn,  they  will 
include  an  angle  which  is  the  supplement  of 
the  given  angle. 


M 


BOOK  I 


73 


116.  The  lines  joining   (in  order)   the  midpoints  of  the  sides  of  a 
quadrilateral  form   a  parallelogram   the   sum   of 

whose  sides  is  equal  to  the  sum  of  the  diagonals  of 
the  quadrilateral.      [Use  142.]  £ 

117.  The    lines   joining   (in   order)    the    mid-    R 
points  of  the  sides  of  a  rectangle  form  a  rhombus. 
[Draw  the  diagonals.] 

118.  If  a  perpendicular  be  erected  at  any  point  in 
the  base  of  an  isosceles  triangle,  meeting  one  leg,  and 
the  other  leg  produced,  another  isosceles  triangle  will 
be  formed. 

[Z    o   and   Z.  S   are     complements  of   =  A   A    and 
B  (?).    Etc.] 

119.  The  difference  between    two  sides    of   a 
triangle  is  less  than  the  third  side. 

120.  The  bisectors  of  two  exterior  angles  of  a 
triangle  and  of  the  interior  angle  at  the  third  ver- 
tex meet  in  a  point. 

121.  The  bisectors  of  the  exterior  angles  of  a 
rectangle  form  a  square. 

122.  If  lines  be  drawn  from  a  pair  of  oppo- 
site vertices  of  a  parallelogram  to  the   mid- 
points of  a  pair  of  opposite  sides,  they  will 
trisect  the  diagonal  joining  the  other  two  ver- 
tices.    [Prove  AECF&EJ&nd  use  142  in  & 
DYCandABX.] 

123.  If  two  medians  of  a  triangle  are  equal, 
the  triangle  is  isosceles. 

[Use  151.  AO  -  OB  (Ax.  3).     Hence  prove  & 
AEO  and  DBO  equal.] 

124.  How  many  sides  has  the  polygon  the  sum 
of  whose  interior  angles   exceeds  the  sum  of  its 
exterior  angles  by  900  °  ? 

125.  If  the  vertex-angle  of  an  isosceles  triangle 
is  twice  the  sum  of  the  base  angles,  any  line  per- 
pendicular to  the  base  forms  with  the  sides  of  the 
given    triangle     (one    .side  to    be    produced)    an 
equilateral  triangle.     [Use  121.] 


O      B 


74 


PLANE   GEOMETRY 


126.  The  lines  bisecting  two  interior  angles  that  a  transversal  makes 
with  one  of  two  parallels  cut  off  equal  segments  on  the  other  parallel  from 
the  point  at  which  the  transversal  meets  it.     [The  &  formed  are  isosceles.] 

127.  The  bisector  of  the  right  angle   of  a 
right  triangle  is  also  the  bisector  of  the  angle 
formed  by  the  median  and  the  altitude  drawn 
from  the  same  vertex. 

To  Prove :  Z  MCS  =  LCS. 

Proof :   Z  A  CS  =  Z  BCS  (?) ;  Z  A  CM  =  Z  BCL  (?) .   Now  use  Ax.  2. 

128.  If  through  the  point  of  intersection  of  the  diagonals  of  a  par- 
allelogram, two  lines  be  drawn   intersecting   a  pair  of  opposite  sides 
(produced  if  necessary),  the  intercepts  on  these  sides  will  be  equal. 

129.  If  ABC  is  a  triangle,  BS  is  the  bisector 
of  Z  ABC,  and  AM  is  parallel  to  BS  meeting  BC 
produced,  at  M,  the  triangle  ABMia  isosceles. 

130.  If  ABC  is  a  triangle    and  BS  is  the 
bisector  of  exterior  Z  ABR  and  AM  is  II  to  BS 
meeting  BC  at  M,  A  ABM  is  isosceles. 

131.  If  A)  B,  C,  and  D  are  points  on  a 
straight  line  and  AB  =  BC,  the  sum  of  the 
perpendiculars  from  A  and  C  to  any  other 
line  through  D  is  double  the  perpendicular 
to  that  line  from  B.    [Use  147 ;  144.] 

132.  If  on  diagonal  BD,  of  square  ABCD,  BE  be 
taken  equal  to  a  side  of  the  square,  and  EP  be  drawn 
perpendicular  to  BD  meeting  AD  at  P,  AP  =  PE  = 
ED.     [Draw  BP.~] 

133.  If  in  A  ABC,  Z  A  is  bisected  by  line  meet- 
ing BC  at  M,  AB>BM  and  AC>  CM.      [Use  108; 
123.] 

134.  It  is  impossible  to  draw  two  straight  lines 
from  the  ends  of  the  base  of  a  triangle  terminating 
in  the  opposite  side,  so  that  they  shall  bisect  each 
other.     [Use  138.] 

135.  If   ABC   is   an  equilateral   triangle   and 
D,  E,  F  are  points  on  the  sides,  such  that  AD  — 

.  BE  =  CF,  triangle  DEF  is  also  equilateral. 
[Prove  the  three  small  &  =.] 


F        C 


BOOK   I 


75 


136.  If  A  BCD  is  a  square  and  E,  F,  <2,  H  are 

points  on  the   sides,  such  that    AK  =  HF  -  <'<;  = 
/)//,  EFGH   is  a  square. 

[First,  prove  EFGH  equilateral ;  then  one  Za  rt.  Z. .] 

137.  If  ABC   is  an  equilateral  triangle  and  each 
side  is  produced  (in  order)  the  same  distance,  so  that 
A  D  =  BE  =  CF,  the  triangle  DEF  is  equilateral. 

138.  If  A  BCD  is  a  square  and  the  sides  be  produced  (in  order)  the 
same  distance,  so  that  AE  =  BF  =  CG.  =  DH,  the  figure  EFGH  will  be 
a  square. 

139.  The  two  lines  joining  the  midpoints  of  the  opposite  sides  of  a 
quadrilateral  bisect  each  other.     [Join  the  4  midpoints  (in  order),  etc.] 

140.  If  two  adjacent  angles  of  a  quadrilateral  are  right  angles,  the 
bisectors  of  the  other  angles  are  perpendicular  to  each  other. 

141.  If  two  opposite  angles  of  a  quadrilateral  are  right  angles,  the 
bisectors  of  the  other  angles  are  parallel. 

142.  Two  isosceles  triangles  are  equal,  if : 

(1)  The  base  and  one  of  the  adjoining  angles  in  the  one  are  equal 
respectively  to  the  base  and  one  of  the  adjoining  angles  in  the  other. 

(2)  A  leg  and  one  of  the  base  angles  in  the  one  are  equal  respectively 
to  a  leg  and  one  of  the  base  angles  in  the  other. 

(3)  The  base  and  vertex-angle  in  one  are  equal  to  the  same  in  the 
other. 

(4)  A  leg  and  vertex-angle  in  one  are  equal  to 
the  same  in  the  other. 

(5)  A  leg  and  the  base  in  one  are  equal  to  the 
same  in  the  other. 

143.  If  upon  the  three  sides  of  any  triangle 
equilateral  triangles  be  constructed  (externally) 
and  a  line  be  drawn  from  each  vertex  of  the  given 
triangle  to  the  farthest  vertex  of  the  opposite  equi- 
lateral triangle,  these  three  lines  will  be  equal. 

Proof:  Z  EAC  =Z  BA F  (?).     Add  to  each  R 
of  these,  /.CAB.    :.  Z  EAB  =  Z  CAF  (?). 
Then  prove  &  EAB  and  CAF  equal. 
Similarly,  A  CA  D  =  A  CEB.     Etc. 

144.  If   two   medians  be  drawn   from   two 
vertices  of  a  triangle  and  produced  their  own 
length  beyond  the  opposite  sides  and  these  ex- 
tremities be  joined  to  the  third  vertex,  these  two 

lines  will  be  equal,  and  in  the  same  straight  line.    [Draw  MP  and  use  142.] 


7t)  PLANK    (JKOMKTRY 

145.  The  median  to  one  side  of  a  triangle  is  less  than  half  the  sum  of 
the  other  two  sides. 

Proof:  (Fig.  of  No.  144.)  Produce  median  BM  its  own  length  to  72, 
draw  RA.     Prove  RA  =  CB.     Prove  RR  <  AB  -f  BC,  etc. 

146.  The  sum  of  the  medians  of  a  triangle  is  less  than  the  sum  of  the 
sides  of  the  triangle.  r>       C 

147.  If  the  diagonals  of  a  trapezoid  / ' 
are  equal,  it  is    isosceles. 

[Draw  DR    and   CS  _L  to  AB ;   and 
prove  rt.  A  ACS  and   BDR   equal,   to    A" 
get  Z.  x  =  Z  ar.] 

148.  If  a  perpendicular  be  drawn  from  each  vertex  of  a  parallelogram 
to  any  line  outside  the  parallelogram,  the  sum  of  those  from  one  pair  of 
opposite  vertices  will  equal  the  sum  of  those  from  the  other  pair. 

[Draw  the  diagonals  ;  use  144.] 

B 

149.  The  sum  of  the  perpendiculars  to  the  legs 
of  an   isosceles  triangle    from    any   point    in   the 
base  equals   the  altitude   upon   one   of  the  legs. 
(That  is,  the  sum  of  the  perpendiculars  from  any 
point  in  the  base  of  an  isosceles  triangle  to  the  equal 
sides  is  constant  for  every  point  of  the  base.) 

[Prove  PE  =  CF  by  184  (1).]  A         P~         ~c 

150.  The  sum  of  the  three  perpendiculars  drawn  from  any  point 
within  an  equilateral  triangle,  to  the  three  sides,  is  constant  for  all 
positions  of  the  point. 

[Draw  a  line  through  this  point  II  to  one  side;  draw  the  altitude  of  the 
A  -L  to  this  line  and  side ;  prove  the  sum  of  the 
three  _ls  =  this  altitude  and  hence,  =  a  constant.] 

151.  The  line  joining  the  midpoints  of  one 
pair  of  opposite  sides  of  a  quadrilateral  and  the 
line  joining  the    midpoints   of    the   diagonals 
bisect  each   other. 


To  Prove  :  LM  and  RS  bisect  each  other. 

152.  If  one  leg  of  a  trapezoid  is  perpendicular  to  the  bases,  the  mid- 
point of  the  other  leg  is  equally  distant  from  the  ends  of  the  first  leg. 
[Draw  the  median.] 

153.  The  median  of  a  trapezoid  bisects  both  the  diagonals. 

154.  The  line  joining  the  midpoints  of  the  diagonals  of  a  trapezoid  is 
a  part  of  the  median,  is  parallel  to  the  bases,  and  is  equal  to  half  their 
difference. 


BOOK   I  77 

155.  If,  in  isosceles  triangle  X  YZ,  A  D  be  drawn  from  A,  the  midpoint 
of  YZ,  perpendicular  to  the  base  XZ,  DZ  =  \  XZ.    [Draw  alt.  from  y.] 

156.  If  ABC  is  an  equilateral  triangle,  the  bisectors  of  angles  B  and  C 
meet  at  D,  DE  be  drawn  parallel  to  AB  meeting  AC  at  E,  and  DF, 
parallel  to  BC  meeting  A  C  at  F,  then  AE  =  ED  =  EF  =  DF=  CF. 

157.  If  A  is  any  point  in  RS  of  triangle  RST,  and  B  is  the  midpoint 
of  RA,  C  the  midpoint  of  AS,  D  the  midpoint  of  ST,  and  E  the  mid- 
point of  TR,  then  BCDE  is  a  parallelogram. 

158.  In  a  trapezoid  one  of  whose  bases  is  ^ 
double  the  other,  the  diagonals  intersect  at  a 

point  two  thirds  of  the  distance  from  each  end 
of  the  longer  base  to  the  opposite  vertex. 
Proof:  Take  M,  the  midpoint  of  AO,  etc. 

159.  If  lines  be  drawn  from  any  vertex  of  a  parallelogram  to  the  mid- 
points of  the  two  opposite  sides,  they  will  divide  the  diagonal  which  they 
intersect,  into  three  equal  parts. 

Proof :   Draw  the  other  diagonal  and  use  151. 

160.  If  the  interior  and  exterior  angles  at  two  vertices  of  a  triangle 
be  bisected,  a  quadrilateral  will  be  formed,  two  of  whose  angles  are  right 
angles  and  the  other  two  are  supplementary. 

161.  The  angle  between  the  bisectors  of  two  angles  of  a  triangle 
equals  half  the  third  angle  plus  a  right  angle. 

162.  If,  in  triangle  ABC,  the  bisectors  of  the  interior  angle  at  B  and 
of  the  exterior  angle  at  C,  meet  at  D,  the  angle  BAG  equals  twice  the 
angle  BDC. 

163.  The  four  bisectors  of  the  angles  of  a  quadrilateral  form  a  second 
quadrilateral  whose  opposite  angles  are  supplementary. 

Proof:  Extend  a  pair  of  opposite  sides  of  the  given  quadrilateral 
to  meet  at  X.  Bisect  the  base  angles  of  the  new  A  formed,  meeting  at  O. 
Then  show  that  Z.  O  equals  one  of  the  A  between  the  given  bisectors, 
and  Z  O  is  supplementary  to  the  angle  opposite. 

164.  The  sum  of  the  angles  at  the  vertices  of  a 
five-pointed  star  (pentagram)  is  equal  to  two  right 
angles. 

Proof:  Draw  interior  pentagon.     Find  number  of 
degrees  in  each  of  its  angles.     Hence  find  Z  A,  etc. 

165.  The  lines  joining  the  midpoints  of  the  op- 
posite sides  of  an  isosceles  trapezoid  are  perpendicular  to  each  other. 


78  PLANE   GEOMETRY 

166.  If  the  opposite  sides  of  a  hexagon  are  equal  and  parallel,  the 
three  diagonals  drawn  between  opposite  vertices  meet  in  a  point. 

167.  In  triangle  ABC,  AD  is  perpendicular  to  BC,  meeting  it  at  D; 
E  is  the  midpoint  of  AB,  and  F  of  A  C;  the  angle  EDF  is  equal  to  the 
angle  EA  F.     [Use  148  ;  55.] 

168.  If  the  diagonals  of  a  quadrilateral  are  equal,  and  also  one  pair 
of  opposite  sides,  two  of  the  four  triangles  into  which  the  quadrilateral 
is  divided  by  the  diagonals  are  isosceles. 

169.  If  angle  A  of  triangle  ABC  equals  three  times  angle  B,  there 
can  be  drawn  a  line  AD  meeting  BC  in  D,  such  that  the  triangles  ABD 
and  A  CD  are  isosceles. 

170.  If  E  is  the  midpoint  of  side  BC  of  parallelogram  ABCD,  AE 
and  BD  meet  at  a  point  two  thirds  the  distance  from  A  to  E  and  from 
Dio  B. 

171.  If  in  triangle  ABC,  in  which  AB  is  not  equal  to  AC,  A  C'  be 
taken  on  AB  (produced  if  necessary)  equal  to  A  C,  and  AB'  be  taken 
on  AC  (produced  if  necessary)  equal  to  AB,  and  B'C'  be  drawn  meeting 
BC  at  D,  then  AD  will  bisect  angle  BA  C. 

Proof:  A  ABC  =  A  A  B'C'  (?)  (52).     /.  their  homologous  parts  are 
equal.     Thus  prove  &BC'D  =  A  B'CD  (54).     Etc. 

172.  If  a  diagonal  of  a  parallelogram  bisects  one  angle,  it  also  bisects 
the  opposite  angle. 

173.  If  a  diagonal  of  a  parallelogram  bisects  one  angle,  the  figure  is 
equilateral. 

174.  Any  line  drawn  through  the  point  of  intersection  of  the  diago- 
nals of  a  parallelogram  divides  the  figure  into  two  equal  trapezoids. 
[See  159.] 

175.  If  AR  bisects  angle  A  of  triangle  A  BC  and  A  T  bisects  the  ex- 
terior angle  at  A,  any  line  parallel  to  AB,  having  its  extremities  in  AR 
and  A  T,  is  bisected  by  A  C. 

176.  If  the  opposite  angles  of  a  quadrilateral  are  equal,  the  figure  is 
a  parallelogram.     [See  165.] 


BOOK   II 
THE  CIRCLE 

191.  A  curved  line  is  a  line  no  part  of  which  is  straight. 

192.  A  circumference  is  a  curved  line  every  point  of  which 
is  equally  distant  from  a  point  within,  called  the  center. 

193.  A  circle  is  a  portion  of  a  plane  bounded  by  a  cir- 
cumference.    [O.] 

194.  A  radius  is  a  straight  line  drawn  from  the  center  to 
the  circumference. 

A  diameter  is  a  straight  line  containing  the  center,  and 
whose  extremities  are  in  the  circumference. 


CIRCUMFERENCE                       SECANT                 CENTRAL   ANGLE  SEMI- 
CIRCLE                                  CHORD                 INSCRIBED   ANGLE  CIRCUMFERENCES 
RADIUS                                TANGENT                              ARC                                  SEMICIRCLES 
DIAMETER  POINT   OF   CONTACT 

A  secant  is  a  straight  line  cutting  the  circumference  in  two 
points. 

A  chord  is  a  straight  line  whose  extremities  are  in  the  cir- 
cumference. 

A  tangent  is  a  straight  line  which  touches  the  circumference 
at  only  one  point,  and  does  not  cut  it,  however  far  it  may  be 
extended.  The  point  at  which  the  line  touches  the  circum- 
ference is  called  the  point  of  contact  or  the  point  of  tangency. 

79 


80  PLANE   GEOMETRY 

195.  A  central  angle  is  an  angle  formed  by  two  radii. 

An  inscribed  angle  is  an  angle  whose  vertex  is  on  the  cir- 
cumference and  whose  sides  are  chords. 

196.  An  arc  is  any  part  of  a  circumference. 

A  semicircumference  is  an  arc  equal  to   half  a   circum- 
ference. 

A  quadrant  is  an  arc  equal  to  one  fourth  of  a  circumference. 
Equal  circles  are  circles  having  equal  radii. 
Concentric  circles  are  circles  having  the  same  center. 


CIRCLES    EXTERNALLY 
TANGENT 


197.  A  sector  is  the  part  of  a  circle  bounded  by  two  radii 
and  their  included  arc. 

A  segment  of  a  circle  is  the  part  of  a  circle  bounded  by  an 
arc  and  its  chord. 

A  semicircle  is  a  segment  bounded  by  a  semicircuraference 
and  its  diameter. 

198.  Two  circles  are  tangent  to  each  other  if  they  are  tan- 
gent to  the  same  line  at  the  same  point.      Circles  may  be 
tangent   to   each   other  internally,  if  the  one  is  within  the 
other,  or  externally,  if  each  is  without  the  other. 

199-  POSTULATE.  A  circumference  can  be  described  about  any 
given  point  as  center  and  with  any  given  line  as  radius. 

Explanatory.  A  circle  is  named  either  by  its  center  or  by 
three  points  on  its  circumference,  as  "  the  O  O,"  or  "  the  O 
ABC." 

The  verb  to  subtend  is  used  in  the  sense  of  "to  cut  off." 


BOOK   TI  rtl 

A  chord  subtends  an  arc.  Hence  an  arc  is  subtended  by 
a  chord. 

An  angle  is  said  .to  intercept  the  arc  between  its  sides. 
Hence  an  arc  is  intercepted  by  an  angle. 

The  hypothesis  is  contained  in  what  constitutes  the  sub- 
ject of  the  principal  verb  of  the  theorem.  (See  59.) 

PRELIMINARY   THEOREMS 

200.  THEOREM.    All  radii  of  the  same  circle  are  equal.    (See  192. ) 

201.  THEOREM.    All  radii  of  equal  circles  are  equal.     (See  196.) 

202.  THEOREM.    The  diameter  of  a  circle  equals  twice  the  radius. 

203.  THEOREM.    All  diameters  of  the  same  or  equal   circles  are 
equal.  (Ax.  3.) 

204.  THEOREM.    The  diameter  of  a  circle  bisects  the  circle  and  the 
circumference. 

Given  :    Any  O  and  a  diameter. 

To  Prove  :  The  segments  formed  are  equal,  that  is,  the 
diameter  bisects  the  circle  and  the  circumference. 

Proof  :  Suppose  one  segment  folded  over  upon  the  other 
segment,  using  the  diameter  as  an  axis.  If  the  arcs  do  not 
coincide,  there  are  points  of  the  circumference  unequally 
distant  from  the  center.  But  this  is  impossible  (?)  (192). 

/.  the  segments  coincide  and  are  equal  (?)  (28).      Q.E.D. 

205.  THEOREM.    With  a  given  point  as  center  and  a  given  line  as 
radius,  it  is  possible  to  describe  only  one  circumference.     (See  192.) 

That  is,  a  circumference  is  determined  if  its  center  and  radius  are 
fixed. 

XOTE.  The  word  "  circle  "  is  frequently  used  in  the  sense  of  "  cir- 
cumference." Thus  one  may  properly  speak  of  drawing  a  circle.  The 
established  definitions  could  not  admit  of  such  an  interpretation  save  as 
custom  makes  it  permissible. 


Ex.     Draw  two  intersecting  circles  and  their  common  chord.     Draw 
two  circles  which  have  no  common  chord.     Draw  figures  to  illustrate  all 
the  nouns  defined  on  the  two  preceding  pages. 
BOBBINS'  PLANE  GEOM.  —  6 


82  PLANE   GEOMETRY 


THEOREMS   AND   DEMONSTRATIONS 

206.   THEOREM.   In  the  same  circle   (or  in  equal  circles)  equal 
central  angles  intercept  equal  arcs. 


Given :    O  O  =  O  C ;   Z  O  =  Z  C. 

To  Prove  :    Arc  AB  =  arc  LM. 

Proof :  Superpose  O  o  upon  the  equal  O  <7,  making  Z  o 
coincide  with  its  equal,  Z  C.  Point  A  will  fall  on  i,  and 
point  B  on  M  (?)  (201). 

Arc  AB  will  coincide  with  arc  LM  (?)  (192). 

.'.AB  =  LM  (?)    (28).  Q.E.D. 

207.  THEOREM.  In  the  same  circle  (or  in  equal  circles)  equal 
arcs  are  intercepted  by  equal  central  angles.  [Converse.] 

Given  :    O  O  =  O  C ;  arc  AB  =  arc  LM. 

To  Prove  :    Z  0  =  Z  C. 

Proof :  Superpose  O  O  upon  the  equal  O  C,  making  the 
centers  coincide  and  point  A  fall  on  point  L.  Then  arc  AB 
will  coincide  with  arc  LM  and  point  B  will  fall  on  point  M. 
(Because  the  arcs  are  =.) 

Hence  OA  will  coincide  with  C£,  and  OB  with  CM  (?)  (39). 

/.Z  0  =  Z  c(?)    (28).  Q.E.D. 

Ex.  1.   Can  arcs  of  unequal  circles  be  made  to  coincide?     Explain. 
Ex.  2.   If  two  sectors  are  equal,  name  the  several  parts  that  must  be 
equal. 


BOOK   II  83 

208.   THEOREM.    In  the  same  circle  (or  in  equal  circles) : 

I.   If  two  central  angles  are  unequal,  the  greater  angle  intercepts  the 
greater  arc. 

II.  If  two  arcs  are  unequal,  the  greater  arc  is  intercepted  by  the 
greater  central  angle.    [Converse.] 


I.    Given :   O  o  =  O  c ;   Z  LCM  >  Z  o. 
To  Prove :    Arc  LM  >  arc  AB. 

Proof :    Superpose  O  O  upon  O  C,  making  sector  AOB  fall 
in  position  of  sector  XCM,  OB  coinciding  with  CM. 
GX  is  within  the  angle  LCM  (Z  LCM  >  Z  O). 
Arc  AB  will  fall  upon  ZJf,  in  the  position  XM  (192). 
.-.  arc  LM  >  arc  XM  (Ax.  5).     That  is,  arc  LM  >  arc  AB. 

Q.E.D. 

II.    Given  :    (?).     To  Prove  :    Z  LCM  >  Z  o. 
Proof :    The  pupil  may  employ  either  superposition,  as  in  I, 
or  the  method  of  exclusion,  as  in  87. 

NOTE.  Unless  otherwise  specified,  the  arc  of  a  chord  always  refers  to 
the  lesser  of  the  two  arcs.  If  two  arcs  (in  the  same  or  equal  circles)  are 
concerned,  it  is  understood  either  that  each  is  less  than  a  semicircumfer- 
ence,  or  each  is  greater. 

Ex.  1.  Two  sectors  are  equal  if  the  radii  and  central  angle  of  one 
are  equal  respectively  to  the  radii  and  central  angle  of  the  other. 

Ex.  2.  If  in  the  figure  of  206,  arcs  AB  and  LM  were  removed,  how 
would  the  remaining  arcs  compare  ? 

Ex.  3.  If  in  the  figure  of  208,  arcs  AB  and  LM  were  removed,  how 
would  the  remaining  arcs  compare  ? 


84  PLANE  GEOMETRY 

209.   THEOREM.    In  the  same  circle   (or  in  equal  circles)  equal 
chords  subtend  equal  arcs. 


Given  :    O  O  =  O  C  ;  chord  AB  =  chord  LM. 
To  Prove  :    Arc  AB  =  arc  LM. 

Proof  :    Draw  the  several  radii  to  the  ends  of  the  chords. 
In  A  OAB  and  CLM,  OA  =  CL,  OB  —  CM  (?)   (201). 
Chord  AB  =  chord  LM  (Hyp.).  .-.  A  OAB  =  A  CLM  (?). 


.'.  arc  AB  =  arc  LM  (?)   (206).  Q.E.D. 

210.  THEOREM.    In  the  same  circle  (or  in  equal  circles)  equal  arcs 
are  subtended  by  equal  chords. 

Given  :    O  0  =  O  C  ;  arc  AB  =  arc  LM. 
To  Prove:    Chord  AB  =  chord  LM. 

Proof:    Draw  the  several  radii  to  the  ends  of  the  chords. 
In    A  OAB    and    CLM,   OA  =  CL,  OB  =  CM  (?)  (201). 
ZO=ZC(?)   (207).      /.A  AOB  =A  CLM  (?). 
.'.  chord  AB  =  chord  LM  (?).  Q.E.D. 

211.  THEOREM.  In  the  same  circle  (or  in  equal  circles)  : 

I.  If  two  chords  are  unequal,  the  greater  chord  subtends  the 
greater  arc. 

II.  If  two  arcs  are  unequal,  the  greater  arc  is  subtended  by  the 
greater  chord. 

I.    Given:    O  O  =  O  C  ;  chord  AB  >  chord  BS. 
To  Prove:    Arc  AB  >  arc  US. 


BOOK   II 


Proof :  Draw  the  several  radii  to  the  ends  of  the  chords. 
In  A  AOB  and  RCS,  AO  =  EC,  BO  ==  SC  (?)  (201). 

Chord  AB  >  chord  RS  (Hyp.).     .-.Z  O  >  Z  C  (?)  (87). 

.*.  arc  AB  >  arc  #s  (?)  (208,  I).  Q.E.D. 

II.    Given:  O  O  =  O  C;  arc  ^45  >  arc  E-8. 

To  Prove  :    Chord  AB  >  chord  RS. 

Proof :  Draw  the  several  radii.  In  A  A  OB  and  RCS,  AO  = 
RC,  BO  =  SC  (?)  (201). 

But  Z  o  >  Z  C  (?)  (208,  II). 

.*.  chord  AB  >  chord  RS  (?)  (86).  Q.E.D. 

212.  THEOREM.  The  diameter  perpendicular  to  a  chord  bisects  the 
chord  and  both  the  subtended  arcs. 

Given  :  Diameter  DR  J_  to  chord 
AB  in  O  O. 

To  Prove:  I.  AM=MB->   II.  AR 

=  RB  and  AD  =  DB. 

Proof:    Draw  radii  to  the   ends 
of  the  chord. 

I.  In  rt.  A  OAM  and  OBM,  OA  = 
OB  (?),  OM=OM  (?). 

.-.  AOAM=  Ao#jf(?). 

Hence,  AM=MB  (?).          Q.E.D. 

II.    Z.AOM  =  ZJ503f  (27).      .'.  AR  =  RB  (?)  (206). 

Also  Z^OD  =  Z  BOD  (?)(49).     .'.AD=DB  (?)(206).  Q.E.D. 


86  PLANE  GEOMETRY 

213.  THEOREM.  The  line  from  the  center  of  a  circle  perpendicular 
to  a  chord  bisects  the  chord  and  its  arc.     Proof :  The  same  as  212. 

214.  THEOREM.   The  perpendicular  bisector  of  a  chord  passes  through 
the  center  of  the  circle.     [0  is  equidistant  from  A  and  B  (?)  (200). 
.-.  it  is  in  the  _L  bisector  of  AB  (?)  (69).] 

215.  THEOREM.   The  line  perpendicular  to  a  radius  at  its  extremity 
is  tangent  to  the  circle. 

Given:  Radius  OA  of 
O  O,  and  RT  _L  to  OA  at  A. 

To  Prove:  RT  tangent 
to  the  circle. 

Proof:  Take  any  point 
P  in  RT  (except  A)  and 
draw  OP. 

OP  >  OA  (?)  (77).  

Hence  P    lies    without  R 
the  O.     (Because  OP  >  radius.) 

That  is,  every  point  (except  A)  of  line  RT  is  without 
the  O. 

Therefore,  RT  is  a  tangent  (Def.  194).  Q.E.D. 

216.  THEOREM.   If  a  line  is  tangent  to  a  circle,  the  radius  drawn 
to  the  point  of  contact  is  perpendicular  to  the  tangent. 

Given  :    RT  tangent  to  O  O  at  A  ;   radius  OA. 
To  Prove :    OA  JL  to  RT. 

Proof:  Every,  point  (except  A)  in  RT  is  without  the  O 
(Def.  194). 

Therefore  a  line  from  O  to  any  point  of  RT  (except  A) 
is  >  OA.  (Because  it  is  >  a  radius.)  That  is,  OA  is  the 
shortest  line  from  o  to  RT.  .*.  OA  is  J_  to  RT  (?)  (77).  Q.E.D. 

217.  COR.   The  perpendicular  to  a  tangent  at  the  point  of  contact 
passes  through  the  center  of  the  circle.    (See  43.) 


BOOK  II 


87 


218.  THEOREM    If  two  circles  are  tangent  to  each  other,  the  line 
joining  their  centers  passes  through  their  point  of  contact. 

Given :   ©  O  and  c 

• 

tangent  to  a  line  at 
AI  and  line  OC. 

To  Prove :  oc 
passes  through  A. 

Proof:  Draw  radii 
OA  and  CA.  OA  is 
_L  to  the  tangent 
and  CA  is  _L  to  the  tangent  (?)  (216). 

.'.  OAC  is  a  st.  line  (?)  (43).     /.  OAC  and  OC  coincide  and 
OC  passes  through  A  (39).  Q.E.D. 

Let  the  pupil  apply  this  proof  if  the  circles  are  tangent  internally. 

219.  THEOREM.  Two  tangents  drawn  to  a  circle  from  an  external 
point  are  equal. 


NOTE.     In  this  theorem  the  word  "  tangent "  signifies  the  distance  be- 
tween the  external  point  and  the  point  of  contact. 

Given  :    O  O  ;  tangents  PA,  PB. 

To  Prove  :    Distance  PA  =  distance  PB. 

Proof  :    Draw  radii  to  the  points  of  contact,  and  join  OP. 
A  GAP  and  OBP  are  rt.  A    (?)  (216). 

In    it.   A  GAP    and    OBP,    OP  =  OP    (?)  ;    OA  =  OB    (?). 
.'.A  GAP  =  A  GBP  (?).      .*.  PA  =  PB    (?).  Q.E.D. 


88 


PLANE   GEOMETRY 


220.    THEOREM.    If  from  an  external  point  tangents  be  drawn  to  a 
circle,  and  radii  be  drawn  to  the  points  of  contact,  the  line  joining  the 
center  and  the  external  point  will  bisect  : 
I.  The  angle  formed  by  the  tangents. 
II.  The  angle  formed  by  the  radii. 

III.  The  chord  joining  the  points  of  contact. 

IV.  The  arc  intercepted  by  the  tangents. 

Proof: 


are  rt.  A  (?). 

They  are  =  .  (Explain.) 


II. 

III.  o    is     equidistant 
from  A  and  B  (?). 

P  is  also  (?)  (219). 

.'.  OP  is  -L  to  AB  at  its  midpoint  (?)  (70). 

IV.  Arc  AX  =  arc  BX  (?)  (206). 


Q.E.D. 


221.  THEOREM.    In  the  same  circle  (or  in  equal  circles)  equal  chords 
are  equally  distant  from  the  center. 

Given  :  O  O  ;  chord  AB  =  chord 
Cl>,  and  distances  OE  and  OF. 

To  Prove  :   OE  =  OF. 

Proof  :   Draw  radii  OA  and  OC. 

In  the  rt.  A  AOE  and  COF, 
AE  =  £  AB  ;  CF  =  |  CD  (213). 

But  AB  =  CD  (Hyp.). 

Hence,  AE  =  CF  (Ax.  3)  ;  and 
AO=CO  (?).  .'.  A  AOE=A  COF  (?).  /.  (XE=  OF  (?).  Q.E.D. 

222.  THEOREM.    In  the  same  circle  (or  in  equal  circles)  chords 
which  are  equally  distant  from  the  center  are  equal. 

Given  :   O  O;  chords  AB  and  CD  ;  distance  OE=  distance  OF. 


BOOK   II 


89 


To  Prove :  chord  AB  =  chord  CD. 

Proof :  Draw  radii  OA  and  OC.  In  rt.  A  AOE  and  COF, 
^10  =  CO  (?);  OF  =  OF  (Hyp.).  /.  A  AOE  =  A  COF  (?). 
.-.  .4F  =  CF  (?).  AB  is  twice  J.F  and  CD  is  twice  CF  (?). 
.  .  AB  =  CD  (Ax.  3).  Q.E.D. 

223.  THEOREM.    In  the  same  circle  (or  in  equal  circles)  if  two 
chords  are  unequal,  the  greater  chord  is  at  the  less  distance  from  the 
center. 

Given :  O  O ;  chord  AB  >  chord 
CD,  and  distances  OE  and  OF. 

To  Prove  :  OE  <  OF. 

Proof  :  Arc  AB  >  arc  CD  (?) 
(211,  I).  Suppose  arc  AH  taken 
on  arc  AB  =  arc  CD.  Draw  chord 
AH.  Draw  OK  _L  to  AH  cutting 
AB  at  I. 

Chord  AH  =  chord  CD  (?)  (210). 

Distance  OK  =  distance  OF  (?)  (221). 

But  OE  <  01  (?)  (77);  and  OI  <  OK  (?)  (Ax.  5). 

.'.  OE  <   OK  (Ax.   11).      /.  OE  <   OF  (Ax.  6).  Q.E.D. 

224.  THEOREM.    In  the  same  circle  (or  in  equal  circles)  if  two 
chords  are  unequally  distant  from  the  center,  the  chord  at  the  less  dis- 
tance is  the  greater. 

Given  :  O  O ;  chords  AB  and  CD ;  distance  OE  <  distance  OF. 
To  Prove :  Chord  AB  >  chord  CD. 

Proof :  It  is  evident  that  chord  AB  <  chord  CD,  or  =  chord 
CD,  or  >  chord  CD.  Proceed  by  the  method  of  exclusion. 

Another  Proof :  On  OF  take  OX—  OE.  At  X  draw  a  chord 
RS  J_  to  OX.  Ch.  ES  is  II  to  ch.  CD  (?).  .-.  arc  RS  >  arc 
CD  (Ax.  5).  .-.  ch.  RS  >  ch.  CD  (?). 

But  ch.  AB  =  c\\.  RS  (?).      .-.  ch.  AB  >  ch.  CD  (Ax.  6). 

Q.E.D. 
225.    COR.    The  diameter  is  longer  than  any  other  chord. 


90 


PLANE   GEOMETRY 


226.  THEOREM.    Through  three  points,  not  in  the  same  straight 
line,  one  circumference  can  be  drawn,  and  only  one. 

Given  :   Points  A  and  B  and  C. 
To  Prove:    I.  (?).     II.  (?). 

Proof :  I.  Draw  lines  AB,  BC, 
AC.  Suppose  their  J_  bisectors,  OZ, 
OX,  OF,  be  drawn.  These  Js  will 
meet  at  a  point  (?)  (85).  Using 
O  as  center  and  OA,  OB,  or  OC  as 
radius,  a  circumference  can  be 
described  through  A,  B,  C  (85). 

II.  These  Js  can  meet  at  only  one  point  (85) ;  that  is, 
there  is  only  one  center.  The  distances  from  O  to  A,  O  to  J?, 
O  to  C,  are  all  equal  (85);  that  is,  there  is  only  one  radius.. 
Therefore  there  is  only  one  circumference  (205).  Q.E.D. 

227.  COR.   A  circumference  can  be  drawn  through  the  vertices  of 
a  triangle,  and  only  one. 

228.  COR.    A  circumference  is  determined  by  three  points. 

229.  COR.   A  circumference  cannot  be  drawn  through  three  points 
which  are  in  the  same  straight  line.     [The  Js  would  be  II.] 

230.  COR.    A  straight  line  can  intersect  a  circumference  in  only 
two  points.  (229.) 

231.  COR.    Two  circumferences  can  intersect  in  only  two  points. 

232.  THEOREM.     If  two  circumferences  intersect,  the  line  joining 
their  centers  is  the  perpendicular  bisector  of  their  common  chord. 

Proof:  Draw  radii  in 
each  O  to  ends  of  AB. 
Point  O  is  equally  dis- 
tant from  A  and  B  (?). 
Point  C  is  equally  dis- 
tant from  A  and  B  (?). 
.*.  OC  is  the  J_  bisector  of 
AB  (?)  (70).  Q.E.D. 


HOOK    II 


233.   THEOREM.    Parallel   lines   intercept  equal  arcs  on  a  circum- 
ference. 

P  M  P 


Given:  A  circle  and  a  pair  of  parallels  intercepting  two 
arcs. 

To  Prove :  The  intercepted  arcs  are  equal. 
There  may  be  three  cases : 

I.  If  the  Us  are  a  tangent  (^45,  tangent  at  P)  and  a  secant 
(CD,  cutting  the  circle  at  E  and  F). 

Proof :  Draw  diameter  to  point  of  contact,  P.  This  di- 
ameter is  _L  to  AB  (216).  PP'  is  also  _L  to  EF  (?)  (95). 
.-.  arc  EP  =  arc  FP  (?)  (212). 

II.  If  the  Us  are  two  tangents  (points  of  contact  being  M 
and  JV). 

Proof :  Suppose  a  secant  be  drawn  II  to  one  of  the  tangents, 
cutting  O  at  R  and  S.  RS  will  be  II  to  other  tangent  (?)  (94). 

/.  arc  MR  =  arc  MS;  arc  UN  =  arc  SN  (proved  in  I). 
Adding,  arc  MEN  =  we  MSN  (Ax.  2). 

III.  If  the  Us  are  two  secants  (one  intersecting  the  O  at  A 
and  B ;  the  other  at  C  and  D) . 

Proof:  Suppose  a  tangent  be  drawn  touching  O  at  P,  II  to 
AB.  This  tangent  will  be  II  to  CD  (?). 

/.  arc  PC  =  arc  PD;  arc  PA  =  arc  PB  (by  I). 
Subtracting,  arc  AC  =  arc  BD  (Ax.  2).  Q.E.D. 


234.  A  polygon  is  inscribed 
in  a  circle,  or  a  circle  is  cir- 
cumscribed about  a  polygon 


if  the  vertices  of  the  poly- 
gon are  in  the  circumference, 
and  its  sides  are  chords. 


t»2  PLANE   GEOMETRY 

A  polygon  is  circumscribed]    ;f  the  gides  of  ^ 

about  a   circle,  or  a  circle  is      are  aii  tangent  to  the  circle, 
inscribed  m  a  polygon 

A  common  tangent  to  two  circles  is  a  line  tangent  to  both 
of  them. 

The  perimeter  of  a  figure  is  the  sum  of  all  its  bounding 

lines. 

EXERCISES  IN  DRAWING  CIRCLES 

1.  Draw  two  unequal  intersecting  circles.     Show  that  the  line  joining 
their  centers  is  less  than  the  sum  of  their  radii. 

2.  Draw  two  circles  externally  (not  tangent)  and  show  that  the  line 
joining  their  centers  is  greater  than  the  sum  of  their  radii. 

3.  Draw  two  circles  tangent  externally.     Discuss  these  lines  similarly. 

4.  Draw  two  circles  tangent  internally.     Discuss  these  lines  similarly. 

5.  Draw  two  circles  so  that  they  can  have  only  one  common  tangent. 

6.  Draw  two  circles  so  that  they  can  have  two  common  tangents. 

7.  Draw  two  circles  so  that  they  can  have  three  common  tangents. 

8.  Draw  two  circles  so  that  they  can  have  four  common  tangents. 

9.  Draw  two  circles  so  that  they  can  have  no  common  tangent. 

StTMMARY 

235.    The  following  summary  of  the  truths  relating  to  magnitudes, 
which  have  been  already  established  in   Book  II,  may  be  helpful  in 
attacking  the  original  work  following. 
I.  Arcs  are  equal  if  they  are : 

(1)  Intercepted  by  equal  central  angles. 

(2)  Subtended  by  equal  chords. 

(3)  Intercepted  by  parallel  lines. 

(4)  Halves  of  the  same  arc,  or  of  equal  arcs. 
IT.   Lines  are  equal  if  they  are : 

(1)  Radii  of  the  same  or  equal  circles. 

(2)  Diameters  of  the  same  or  equal  circles. 

(3)  Chords  which  subtend  equal  arcs. 

(4)  Chords  which  are  equally  distant  from  the  center. 

(5)  Tangents  to  one  circle  from  the  same  point. 

III.   Unequal  arcs  and  unequal  chords  have  like  relations. 
[See  208;  211;  223;  224.] 


BOOK    II 


ORIGINAL   EXERCISES 

1.  A  diameter  bisecting  a  chord  is  perpendicular  to  the 
chord  and  bisects  the  subtended  arcs.     [Use  70.] 

2.  A  diameter  bisecting  an  arc  is  the    perpendicular 
bisector  of  the  chord  of  the  arc.     [Draw  AR  and  BR.~\ 

3.  A  line  bisecting  a  chord  and  its  arc  is  perpendicular 
to  the  chord. 

4.  The   perpendicular  bisectors  of  the  sides  of  an  in- 
scribed polygon  meet  at  a  common  point. 

5.  A  line  joining  the  midpoints  of  two  parallel  chords 
passes  through  the  center  of  the  circle. 

[Suppose  diam.  drawn  _L  to  AB\  this  will  be  ±  to  CD.  Etc.] 

6.  The  perpendiculars  to  the  sides  of  a  circumscribed 
polygon  at  the  points  of  contact  meet  at  a  common  point. 
[Use  217.] 

7.  The  bisector  of  the  angle  between  two  tangents  to  a  circle  passes 
through  the  center.     [Use  80.] 

8.  The  bisectors  of  the  angles  of  a  circumscribed  polygon  all  meet 
at  a  common  point. 

9.  Tangents  drawn  at  the  extremities  of  a  diameter  are  parallel. 

10.  In  the  figure  of  220,  prove  Z  APO-Z.  ABO. 

11.  In  the  same  figure,  prove  Z  PA  B  =  Z.  FOB.  [Use  48.] 

12.  If  two   circles  are   concentric,   all  chords   of  the 
greater,  which  are  tangent  to  the  less,  are  equal. 

[Draw  radii  to  points  of  contact.     Use  216  ;  222.] 

13.  Prove  225  by  drawing  radii  to  the  ends  of  the  chord. 

14.  An  inscribed  trapezoid  is  isosceles.     [Use  233.] 

15.  The  line  joining  the  points  of  contact  of  two  parallel  tangents 
passes  through  the  center.    [Draw  radii  to  points  of  contact.     Etc.] 

16.  A  chord  is  parallel  to  the  tangent  at  the  midpoint 
of  its  subtended  arc.     [Draw  radii  to  point  of  contact 
and  to  the  ends  of  the  chord.     Also  draw  chords  of  the 
halves  of  the  given  arc.] 

17.  The  sum  of  one  pair  of  opposite  sides  of  a  cir- 
cumscribed  quadrilateral  is  equal  to  the   sum   of  the 
other  pair.     [Use  219  four  times,  keeping  R  and  T  on 
the  same  side  of  the  equations.] 


94  PLANE   GEOMETRY 

18.  A  circumscribed  parallelogram  is  equilateral. 

19.  A  circumscribed  rectangle  is  a  square. 

20.  If  two  circles  are  concentric  and  a  secant  cuts  them  both,  the  por- 
tions of  the  secant  intercepted  between  the  circumferences   are    equal. 
[Use  212.] 

21.  Of  all  secants  that  can  be  drawn  to  a  circumference  from  a  fixed 
external  point,  the  longest  passes  through  the  center. 

To  Prove :  PB  >  PE.  rv 

22.  The  shortest  line  from  an  external  point  to 
a  circumference  is  that  which,  if  produced,  would 
pass  through  the  center. 

To  Prove :  PA  <  PD.     Draw  CD. 

23.  If  two  equal  secants  be  drawn  to  a  circle  from  an  external  point, 
their  chord  segments  will  be  equal.      [Draw  OA, 

OP,  OC,   OB,   OD.     Prove    &  POD    and    POB  c^ — >^n 

equal;  then  A  COD  and  A  OB  are  equal.]  ^ "*S^"--/S> ~ 

24.  In   No.  23   prove  the  external  segments  "VVZ   \ 


25.  State  and  prove  the  converse  of  No.  23. 

26.  If  two  equal  secants  be  drawn  to  a  circle  from  an  external  point, 
they  will  be  equally  distant  from  the  center. 

27.  If  two  equal  chords  intersect  on  the  circumference,  the  radius 
drawn  to  their  point  of  intersection  bisects  their  angle. 

[Draw  radii  to  the  other  extremities  of  the  chords.] 

28.  Any  two  parallel  chords  drawn  through  the  ends  of  a  diameter 
are  equal. 

29.  If  a  circle  be  inscribed  in  a  right  triangle, 
the  sum  of  the  diameter  and  hypotenuse  will  be 
equal  to  the  sum  of  the  legs. 

[Draw  radii  OR,  OS;  ROSC  is  a  square  (?)  ; 
then  prove  diameter  +  AB  =  A  C  +  BC.'] 

30.  The  shortest  chord  that  can  be  drawn  through  a  given  point 
within  a  circle  is  perpendicular  to  the  diameter  through 

the  point. 

Given:   P,  the  point;  BOC  the  diam. ;  LS  ±  to  BC   G' 
at  P;  OR  any  other  chord  through  P.     To  Prove:  (?). 
Proof :  Draw  OA  _L  to  GR.     Etc. 


ROOK    TT 


95 


31.  What,  is  the  longest  chord  that  can  be  drawn  through  a  given  point 
within  a  circle? 

32.  If  the  line  joining  the  point  of  intersection  of  two  chords  and  the 
center  bisects  the  angles  formed  by  the  chords,  they  are 

equal.     [Draw  .&  OE  and  OF  and  prove  them  = .     Etc.] 

33.  AB  and  AC  are  two  tangents  from  A ;  in  the  less 
arc  EC  a  point  D  is  taken  and  a  tangent  drawn    at  Z>, 
meeting  AB  at  E  and  AC  at  F\  AE  +  EF  +  AFequals  a 
constant  for  all  positions  of  D  in  arc  BC. 

[Prove  this  sum  =  AB  +  BC.] 

34.  The  radius  of  the   circle   inscribed  in   an  equi- 
lateral triangle  is  half  the  radius  of  circle  circumscribed 
about  it.     [Use  152.] 

35.  If  the  inscribed  and   circumscribed  circles  of  a 
triangle  are  concentric,  the  triangle  is  equilateral. 

36.  If  two  parallel  tangents  meet  a  third  tangent 
and  lines  be  drawn  from   the  points  of  intersection 
to  the  center,  they  will  be  perpendicular. 

37.  Tangents  drawn  to  two  tangent  circles  from 
any  point  in  their  common  interior  tangent  are  equal. 

38.  The  common  interior  tangent  of  two  tangent 
circles  bisects  their  common  exterior  tangent. 

39.  Do  the  theorems  of  No.  37  and  No.  38  apply 
if  the  circles  are  tangent  internally  ?    If  so,  prove. 

40.  In  the  adjoining  figure  if   AE  and  AD  are 
secants,  A  E  passing  through  the  center,  and  the  ex- 
ternal  part   of  AD  is  equal  to  a  radius,   the  angle 
DCE  =  3Z^. 

[Draw  BC.    ZDBC  =  ext.  Z  of  &ABC  = 
=  ZD  (explain).     Z  DCE  =  an  ext.  Z,  etc.] 

41.  If  perpendiculars  be  drawn  upon  a  tangent 
from  the  ends  of  any  diameter : 

(1)  The  point  of  tangency  will  bisect  the  line 
between  the  feet   of  the   perpendiculars. 

[Draw  CP.     Use  144.]  O         P 

(2)  The  sum  of  the  perpendiculars  will  equal  the  diameter. 

(  3)  The  center  will  be  equally  distant  from  the  feet  of  the  perpendic- 
ulars.    [Use  67.] 


9<)  PLANE   GEOMETRY 

42.    The  two  common  interior  tangents  of  two  circles  are  equal. 


C 

43.  The  common  exterior  tangents  to  two  circles  are  equal. 
[Produce  them  to  intersection.] 

44.  In  the  above  figure,  prove  that  RH  =  SF. 

Proof:  AR  +  RB  =  CS  +  SD  ;  /.  AR  +  (RH  +  FIF)  =  (SF  + 
HF)  +  SD. 

.'.  RH  +  RH  +  HF  =  SF  +  HF  +  SF ;  /.  2  RH  =  2  SF,  etc. 
Give  reasons  and  explain. 

45.  The  common  exterior  tangents  to  two  circles  intercept,  on  a  com- 
mon interior  tangent  (produced),  a  line  equal   to   a   common  exterior 
tangent.     To  Prove  :    RS  =  A  B. 

46.  Prove  that  in  the  figure  of  No.  42  the  line  joining  the  centers 
will  contain  0  and  0'. 

47.  Prove  that  in  the  figure  of  No.  42  if.  chords  AC  and  BD  are 
drawn,  they  are  parallel. 

48.  If  a  circle  be  described  upon  the  hypotenuse  of  a  right  triangle  as 
a  diameter,  it  will  contain  the  vertex  of  the  right  angle  (142). 

49.  The  median  of  a  trapezoid  circumscribed  about  a  circle  equals 
one  fourth  the  perimeter  of  the  trapezoid. 

50.  If  the  extremities  of  two  perpendicular  diameters  be  joined  (in 
order),  the  quadrilateral  thus  formed  will  be  a  square. 

51.  If  any  number  of  parallel  chords  of  a  circle  be  drawn,  their  mid- 
points will  be  in  the  same  straight  line. 

52.  State  and  prove  the  converse  of  No.  35. 

53.  The  line  joining  the  center  of  a  circle  to  the  point 
of  intersection  of  two  equal  chords  bisects  the  angle  formed 
by  the  chords. 


BOOK  IT  97 

KINDS  OF   QUANTITIES  — MEASUREMENT 

236.  A  ratio  is  the  quotient  of  one  quantity  divided  by 
another  —  both  being  of  the  same  kind. 

237.  To  measure  a  quantity  is  to  find  the  number  of  times 
it   contains  another  quantity  of  the  same  kind,    called  the 
unit.     This  number  is  the  ratio  of  the  quantity  to  the  unit. 

238.  Two  quantities  are   called   commensurable   if   there 
exists  a  common  unit  of  measure  which  is  contained  in  each 
a  whole  (integral)  number  of  times. 

Two  quantities  are  called  incommensurable  if  there  does 
not  exist  a  common  unit  of  measure  which  is  contained  in 
each  a  whole  number  of  times. 

Thus  :   $17  and  $35  are  commensurable,  but  $  17  and  $  V35  are  not. 
Two  lines  18^  ft.  and  13  yd.  are  commensurable,  but  18£  in.  and 
\/T3  mi.  are~/io/. 

239.  A  constant  quantity  is  a  quantity  whose  value  does 
not  change  (during  a  discussion).      A  constant  may  have 
only  one  value. 

A  variable  is  a  quantity  whose  value  is  changing.  A 
variable  may  have  an  unlimited  number  of  values. 

240.  The  limit  of  a  variable  is  a  constant,  to  which  the 
variable  cannot  be  equal,  but  from  which  the  variable   can 
be  made  to  differ  by  less  than  any  mentionable  quantity. 

241.  Illustrative.     The  ratio  of  15  yd.  to  25  yd.  is  written  either  £f 
or  15-4-25  and  is  equal  to  three  fifths.     If  we  state  that  a  son  is  two 
thirds  as  old  as  his  father,  we  mean  that  the  son's  age  divided  by  the 
father's  equals  two  thirds.     A  ratio  is  a  fraction. 

The  statement  that  a  certain  distance  is  400  yd.  signifies  that  the  unit 
(the  yard),  if  applied  to  this  distance,  will  be  contained  exactly  400  times. 

Are  $7.50  and  $3.58  commensurable  if  the  unit  is  «fl?  Idime? 
1  cent? 

Are  10  ft.  and  Vl9  ft.  commensurable  ? 
BOBBINS'  PLANE  GEOM.  —  7 


98  PLANE   GEOMETRY 

The  height  of  a  steeple  is  a  constant ;  the  length  of  its  shadow  made 
by  the  sun  is  a  variable.  Our  ages  are  variables.  The  length  of  a 
standard  yard,  mile,  or  meter,  etc.,  is  a  constant.  The  height  of  a  grow- 
ing plant  or  child  is  a  variable. 

The  limit  of  a  variable  may  be  illustrated  by  considering  a  right  tri- 
angle ABC,  and  supposing  the  vertex  A 
to  move  farther  and  farther  from  the 
vertex  of  the  right  angle.  It  is  evident 
that  the  hypotenuse  will  become  longer, 
that  A  C  will  increase,  but  EC  will  re- 
main the  same  length.  The  angle  A. 
must  decrease,  the  angle  B  must  in- 
crease, but  the  angle  C  remains  con- 
stantly a  right  angle.  If  we  carry  vertex  A  toward  the  left  indefinitely, 
the  Z  A  will  become  less  and  less  but  cannot  become  zero.  [Because, 
then  there  could  be  no  A.] 

Hence,  the  limit  of  the  decreasing  Z  A  is  zero  (240). 

Likewise,  the  Z  .B  will  become  larger  and  larger  but  cannot  become 
equal  to  a  right  angle.  [Because,  then  two  sides  of  the  triangle  would  be 
parallel,  which  is  impossible.]  But  it  may  be  made  as  nearly  equal  to  a 
right  angle  as  we  choose. 

Hence,  the  limit  of  Z  B  is  a  right  angle  (240). 

To  these  limits  we  cannot  make  the  variables  equal,  but  from  these 
limits  we  can  make  them  differ  by  less  than  any  mentionable  angle,  how- 
ever small. 

The  following  supplies  another  illustration  of  the  limit  of  a  variable. 
The  sum  of  the  series  l+i  +  J+|  +  TV  +  A  +  B?  +  T£*  +  etc-  etc.,  will 
always  be  less  than  2,  no  matter  how  many  terms  are  collected.  But  by 
taking  more  and  more  terms  we  can  make  the  actual  difference  between 
this  sum  and  2  less  than  any  conceivable  fraction,  however  small. 
Hence,  2  is  the  limit  of  the  sum  of  the  series.  The  limit  is  not  3  nor  4, 
because  the  difference  between  the  sum  and  3  cannot  be  made  less  than 
any  assigned  fraction.  Neither  is  the  limit  1$.  (Why  not?)  Similarly, 
the  limit  of  the  value  of  .333333 ad  infinitum  is  \. 

Certain  variables  actually  become  equal  to  a  fixed  magnitude;  but 
this  fixed  magnitude  is  not  a  limit  (240).  Thus  the  length  of  the 
shadow  of  a  tower  really  becomes  equal  to  a  fixed  distance  (at  noon). 
A  man's  age  really  attains  to  a  definite  number  of  years  and  then  ceases 
to  vary  (at  death). 

Such  variables  have  no  limit  in  the  mathematical  sense  of  that 
word. 


BOOK  II  99 

Hence  :  If  a  variable  approaches  a  constant,  and  the 
difference  between  the  two  can  be  made  indefinitely  small 
but  they  cannot  become  equal,  the  constant  is  the  limit  of 
the  variable.  This  is  merely  another  definition  of  a  limit. 

242.  THEOREM  OF  LIMITS.    If  two  variables  are  always  equal  and 
each  approaches  a  limit,  their  limits  are  equal. 

Given:  Two  variables  v  and  v';  v  always  —vf-t  also  v  ap- 
proaching the  limit  1;  v'  approaching  the  limit  I' . 

To  Prove :  1=1'. 

Proof:  v  is  always  =vr  (Hyp.).  Hence  they  may  be  con- 
sidered as  a  single  variable.  Now  a  single  variable  can 
approach  only  one  limit  (240).  Hence,  1=  I'.  Q.E.D. 

NOTE.  In  order  to  make  use  of  this  theorem,  one  must  have,  first,  two 
variables;  second,  these  must  be  always  equal;  third,  they  must  each  ap- 
proach a  limit.  Then,  the  limits  are  equal. 

243.  (1)   Algebraic  principles  concerning  variables. 

If  v  is  a  variable  and  &  is  a  constant : 

I.    v  -f  k  is  a  variable.  IV.    kv  is  a  variable. 

II.    v  —  k  is  a  variable.  V.    -  is  a  variable. 

k 

III.    k  ±  v  is  a  variable.  VI.    -  is  a  variable. 

v 
These  six  statements  are  obvious. 

(2)   Algebraic  principles  concerning  limits. 

If  v  is  a  variable  whose  limit  is  Z,  and  &  is  a  constant : 
I.    v  ±  k  will  approach  I  ±  k  as  a  limit. 
II.    k  ±  v  will  approach  k  ±  I  as  a  limit. 

III.  kv  will  approach  Id  as  a  limit. 

IV.  ^  will  approach  -  as  a  limit. 

tC  /C 

k  k 

V.  -  will  approach  -  as  a  limit. 

XOTK.  In  these  principles  as  applied  to  Plane  Geometry,  a  variable 
is  not  added  to,  nor  subtracted  from,  nor  multiplied  by,  nor  divided  by 
another  variable.  These  operations  present  little  difficulty,  however. 


100  PLANE   GEOMETRY 

Proofs  :  I.   v  cannot  =  /  (240).     .-.  v  ±  k  cannot  =  I  ±  k. 
Also,  v  —  I  approaches  zero  (240). 

.-.  (v  ±k}  —  (l  ±  k)  approaches  zero.     (Because  it  reduces  to  v  —  I.) 
Hence,  v  ±  k  approaches  /  ±  k  (240). 
II.    Demonstrated  similarly. 

III.   If  ko  =  kl,  then  v  =  l   (Ax.  3).     But  this  is  impossible  (240). 
.•.  kv  cannot  =  kl. 

Also  v  —  I  approaches  zero  (240). 
.-.  k(v  —  1)  or  kv  —  kl  approaches  zero. 
Therefore  kv  approaches  kl  (240). 
IV  and  V.   Demonstrated  similarly. 

244.    THEOREM.    In  the  same  circle  (or  in  equal  circles)  the  ratio 
of  two  central  angles  is  equal  to  the  ratio  of  their  intercepted  arcs. 


Given :  O  o  =  O  C  ;  central  A  O  and  C;  arcs  AB  and  XY. 

To  Prove:   ^=arc^. 
Z  C      arc  XT 

Proof :  I.  If  the  arcs  are  commensurable.  There  exists  a 
common  unit  of  measure  of  AB  and  XY  (238).  Suppose 
this  unit,  when  applied  to  the  arcs,  is  contained  5  times  in 

AB  and  7  times  in  XY.     .-.  arc  AB  =  j>  (Ax.  3).     Draw  radii 

HYCXY     7 

to    the    several    points   of    division    of   the   arcs.      Z  O    is 
divided  into  5  parts,  Z  c  into  7  parts  ;  all  of  these  twelve 

parts  are  equal  (?)  (207).     .-.—  =  |  (Ax.  3). 

£.  c      i 

Z  O  _  arc  AB    ,.       -,, 
'  '  Z  c  ~  arc  XY  ^    X'     }'  Q-E.D. 


BOOK   II  101 

II.  If  the  arcs  are  incommensurable.  There  does  not 
exist  a  common  unit  (238).  Suppose  arc  AB  divided  into 
equal  parts  (any  number  of  them).  Apply  one  of  these  as 
a  unit  of  measure  to  arc  XY.  There  will  be  a  remainder 
PY  left  over.  (Because  AB  and  XT  are  incommensurable.) 


Draw  CP.     Now     Z  °    =  arc  AB.    (The  case  of  commen- 
Z  XCP     arc  XP 

surable  arcs.) 

Indefinitely  increase   the  number   of  subdivisions  of  arc 
AB.     Then  each  part,  that  is,  our  unit  or  divisor,  will  be 
indefinitely  decreased.     Hence  PF,  the  remainder,  will  be  in- 
definitely decreased.    (Because  the  remainder  <  the  divisor.) 
That  is,  arc  PY  will  approach  zero  as  a  limit 
and  Z  PCY  will  approach  zero  as  a  limit. 

.'.  arc  XP  will  approach  arc  XY  as  a  limit  (240) 
and  Z  XCP  will  approach  Z  XCY  as  a  limit  (240). 

will  approach  — as  a  limit  (243) 


Z  XCP  Z  XC 

'I-,  i_  a 

will  approach 
a 

Z  O         arc  AB 


j  arc  AB      'I-,  i_  arc  AB  v     .,   ^o/iox 

and  -         -  will  approach  --  as  a  limit  (243). 
arc  XP  arc  XY 


Z  XCY     arc  XY 


Ex.  How  many  degrees  are  there  in  a  central  angle  which  intercepts 
I  of  the  circumference?  \  of  the  circumference?  ^  of  the  circumfer- 
ence ?  T5y  of  the  circumference  ? 


102 


PLANE   GEOMETRY 


245.    THEOREM.    A  central  angle  is  measured  by  its  intercepted  arc. 

Given:    O  o;  Z.AOY;  arc  AY. 

To  Prove:  Z  AOY  is  measured 
by  the  arc  A  F,  that  is,  they  contain 
the  same  number  of  units. 

Proof :  The  sum  of  all  the  A 
about  0=4  rt.  A  =  360°  (?)  (47). 

If  the  circumference  of  this  O 
be  divided  into  360  equal  parts  and 
radii  be  drawn  to  the  several  points 
of  division,  there  will  be  360  equal  central  A  (207). 

Each  of  these  360  central  angles  will  be  a  degree  of  'angle (21). 

Suppose  we  call  each  of  the  360  equal  arcs,  a  degree  of  arc. 
Take  Z  AOT,  one  of  these  degrees  of  angle,  and  arc  AT,  one 


of  the  degrees  of  arc.     Then, 


Z  ^4OF_arc  AY 
~arc  AT 


(?)  (244). 


But  =  Z^ior-5-a  unit  of   angle  =  the    number  of 

Z  AOT 

units  in  /.AOY  (237). 


And 


=  arc   AY  -5-  a   unit   of   arc  =  the   number  of 


arc  AT 
units  in  arc  AY  (237). 

Hence,  the  number  of  units  in  Z  AOY  =  the  number  of  units 
in  arc  AY  (Ax.  1). 

That  is,  Z  AOY  is  measured  by  arc  AY.  Q.E.D. 

246.  COR.    A   central  right  angle  intercepts  a  quadrant   of  arc. 
(Because  each  contains  90  units.) 

—^  -v 

247.  COR.    A  right  angle  is  measured  by  half 
a  semicircumf  erence,  that  is,  by  a  quadrant. 

248.  An  angle  is  inscribed  in  a  segment 
if  its  vertex  is  on  the  arc  and  its  sides  are 
drawn  to  the  ends  of  the  arc  of  the  segment. 

Thus  ABCD  is  a  segment  and  Z  ABD  is  inscribed  in  it, 


BOOK  II 


103 


249.   THEOREM.  An  inscribed  angle  is  measured  by  half  its  inter- 
cepted arc. 

Given  :    O  O  ;  inscribed  Z  A  ;  arc  CD. 

To  Prove :    Z  A  is  measured  by  J  arc  CD. 

Proof:  I.  If  one  side  of  the  Z  is  a 
diameter.  Draw  radius  CO.  A  AOC  is 
isosceles  (?).  .-.  Z  A  =  Z  C  (?). 

Z  COD  =  ZA  +  Z  C  (?)  (108). 

.-.  Z  COD  =  Z^i-r-^^  =  2Z^i  (Ax.  6). 

That  is,  Z  A  =  %  Z  COD  (Ax.  3). 

But  Z  COD  is  measured  by  arc  CD  (?)  (245). 

.*.  J  Z  COD  is  measured  by  ^  arc   CD  (Ax.  8). 

Therefore,  Z  A  is  measured  by  J  arc  CD  (Ax.  6). 
A  A 


X  X 

II.    If  the  center  is  within  the  angle.     Draw  diameter  AX. 
Z.CAX  is  measured  by  J  arc  CX  (I). 

is  measured  by  J  arc  DX  (I).     Adding, 

is  measured  by 


arc  CD  (Ax.  2). 
III.    If  the  center  is  without  the  angle.    Draw  diameter  AX. 
/.CAX  is  measured  by  ^  arc  CX  (I). 
ZD-4JT  is  measured  by  |   arc  DJT  (I).     Subtracting, 

Z  C.4D  is  measured  by  -|  arc  CD    (Ax.  2).  Q.E.D. 

NOTE.     It  is  evident  that  angles  measured  by  |  the  same  arc  are  equal. 


Ex.  1 .   In  the  figure  of  249,  if  arc  CD  is  56°,  how  many  degrees  are 
there  in  angle  A  ?    If  arc  CD  is  108°,  how  many  degrees  are  in  angle  A  V 
Ex'.  2.   If  Z.A  contains  35°,  how  many  degrees  are  there  in  arc  C/)? 


104 


PLANE   GEOMETRY 


250.  THEOREM.   All  angles  inscribed  in  the 
same  segment  are  equal. 

Given  :  The  several  A  A  inscribed  in 
segment  BAG. 

To  Prove :  These  angles  all  equal. 

Proof :  Each  Z  BAG  is  measured  by 
1  arc  BG  (?)  (249). 

.-.  they  are  equal.  (Because  they  are 
measured  by  half  the  same  arc.)  Q.E.D. 

251.  COR.    All  angles  inscribed  in  a  semi- 
circle are  right  angles. 

Proof  :  Each  is  measured  by  half  of  a 
semicircumference  (?)  (249). 

.-.  each  is  a  rt.  Z  (?)  (247).         Q.E.D. 


252.   THEOREM.   The  angle  formed  by  a  tangent  and  a  chord  is 
measured  by  half  the  intercepted  arc. 

Given  :  Tangent  TN ; 
chord  PA  ;  Z  TPA  ;  arc 
PA. 

To  Prove  :  Z  TPA  is 
measured  by  ^  arc  P^l. 

Proof  :  Draw  diameter 
PX  to  point  of  contact. 
Z  TPX  is  a  rt.  Z  (?)  (216)  ; 
arc  PAX  is  a  semicircum- 
ference (?)  (204). 

Z  TPX  is  measured  by  %  arc  PAX  (?)  (247). 

^  APX  is  measured  by  j.  arc  AX  (?)  (249).  Subtracting, 

Z  TPA   is  measured  by  £  arc     PA    (Ax.  2).  Q.E.D. 

Similarly,  Z  NPA  is  measured  by  ^  arc  PBA. 

(Use  Z  NPX,  and  add.) 


BOOK    II 


105 


(233). 


253.  THEOREM.   The    angle  formed   by   two   chords    intersecting 
within  the  circumference  is  measured  by  half  the  sum  of  the  inter- 
cepted arcs.     (The     arcs    are    those 

intercepted    by    the    given    angle 
and  by  its  vertical  angle.) 

Given  :  Chords  AB  and  CD  in- 
tersecting at  P;  Z  APC;  arcs  AC 
and  DB. 

To  Prove  :  Z  APC  is  measured 
by  ^  (arc  AC+  arc  DB). 

Proof :  Suppose  CX  drawn  through  C  \\  to  AB. 

Now  Z  c  is  measured  by  ^  arc  DX  (?)  (249). 

That  is,  Z  C  is  measured  by  ^  (arc  BX  +  arc 

But  Z  C  =  Z  APC  (?)  (97)  and  arc  BX  =  arc  ^ 

.-.  /.APC  is  measured  by  |-  (arc  .4(7  +  arc  DB)  (?)  (Ax.  6). 

Q.E.D. 

NOTE.     This  theorem  may  be  proven  by  drawing  chord  A  D.     Then 
/.APC  is  an  ext.  Z  of  A  ADP  and  =  £A  +  /.D  (?).     [Use  249.] 

254.  THEOREM.    The  angle  formed  by  two  tangents  is  measured 
by  half  the  difference  of  the  intercepted  arcs. 

Given :  The  two  tan- 
gents AC  and  AB\  Z-4; 
arcs  CMB  and  CNB. 

To  Prove  : 

Z  A  is  measured  by 
J(arc  CMB  —  arc  CNB). 

Proof  :    Suppose 
drawn  ||   to  ^LB. 

Now  Z  Z)CX  is  measured  by  J  arc  CX  (?)  (252). 

That  is,  ZDCX  is  measured  by  £  (arc  CJtfB  —  arc 

But  Z  /)CX  =Z  ,4  (?)  (98)  ;   arc  BX  =  arc  CNB  (?). 

.'. Z  ^  is  measured  by  J  (arc  OJfl?  —  arc  CNB)  (Ax. 6).  Q.E.D. 


106 


PLANE   GEOMETRY 


255.  THEOREM.  The  angle  formed  by  two  secants  which  intersect 
without  the  circumference  is  measured  by  half  the  difference  of  the  in- 
tercepted arcs. 

Given:   (?).     To  Prove:  (?). 
Proof:    Suppose   BX  drawn. 

Where?  How? 

Z  CB x  is  measured  by 
J  arc  CX  (?). 

That  is,  by  |(arc  CE 
—  arc  EX). 

But  Z  CBX  =  Z  A  (?) ;  arc  EX  =  arc  BD  (?). 

.-.  ZA  is  meas.  by  J  (arc  OE  — arcJ5D)  (Ax.  6).        Q.E.D. 

256.  THEOREM.  The  angle  formed  by  a  tangent  and  a  secant  which 
intersect  without  the  circumference,  is  measured  by  half  the  difference 
of  the  intercepted  arcs.  c 

Given:  (?). 

To  Prove :  (?). 

Proof:   Suppose  BX  drawn 
etc. 

Z  CBX  is  measured  by  J 
arc  BX  (?)  (252). 

That  is,  by  J  (arc  B XE  —    A 
arc  EX).     Etc. 


NOTE.  The  theorem  of  254  maybe  proven  by  drawing  chord  BC. 
Then  Z  DCB  =  Z  A  +  Z  OB.4  (?)  ;  or,  Z  ^  =  Z  DOB  -  Z  C.B.4  (Ax.  2). 

Z  Z)<7£  is  measured  by  \  arc  CATS  (?)  and  Z  C5.4  is  measured  by  \ 
arc  CW£  (?).  Hence,  Z  ^  is  measured  by  |  (arc  CMB  -  arc  CNB)  (?). 


Ex.  1.    Prove  the  theorem  of  253  for  angle  A  PD  by  drawing  chord  A  C. 

Ex.  2.  Prove  the  theorem  of  255  by  drawing  chord  CD.  Again,  by 
drawing  chord  BE. 

Ex.  3.  Prove  the  theorem  of  256  by  drawing  chord  BD.  Again,  by 
chord  BE. 


BOOK  II  107 

ORIGINAL   EXERCISES 

1.  If  an  inscribed  angle  contains  20°,  how  many  degrees  are  there  in 
its  intercepted  arc  ?     How  many  degrees  are  there  in  the  central  angle 
which  intercepts  the  same  arc  ? 

2.  A  chord  subtends  an  arc  of  74°.     How  many  degrees  are  there 
in  the  angle  between  the  chord  and  a  tangent  at  one  of  its  ends? 

3.  How  many  degrees  are  there  in  an  angle  inscribed  in  a  segment 
whose  arc  contains  210°?  in  a  segment  whose  arc  contains  110°?   40°? 

4.  Two  intersecting  chords  intercept  opposite  arcs  of  28°  and  80°. 
How  many  degrees  are  there  in  the  angle  formed  by  the  chords  ? 

5.  The  angle  between  a  tangent  and  a  chord  contains  27°.  How  many 
degrees  are  there  in  the  intercepted  arc  ? 

6.  The  angle  between  two  chords  is  30° ;  one  of  the  arcs  intercepted 
is  70° ;  find  the  other  arc.     [Denote  the  arc  by  z.] 

7.  If  in   figure  of  252,  arc  AP  contains  124°,    how  many  degrees 
are  there  in  £  A  PX  ?    in  Z  NPA  ? 

8.  If  in  figure  of  253,  arc  AC  is  85°,  Z  A  PC  is  47°,  find  arc  DB. 

9.  If  the  arcs  intercepted  by  two  tangents  contain  80°  and  280°,  find 
the  angle  formed  by  the  tangents. 

10.  If  the  arcs  intercepted  by  two  secants  contain  35°  and  185°,  find 
the  angle  formed  by  the  secants. 

11.  If  in  figure  of  254,  arc  CB  is  135°,  find  the  angle  A. 

12.  If  in  figure  of  255,  angle  A  =  42°  and  arc  ED  =  70°,  find  arc  CE. 

13.  If  in  figure  of  256,  angle  A  =  18°,  arc  BXE  =  190°,  find  arc  BD. 

14.  If  the  angle  between  two  tangents  is  80°,  find  the  number  of 
degrees  in  each  intercepted  arc.     [Denote  the  arcs  by  x  and  360° — x.~] 

15.  The  circumference  of   a  circle  is  divided  into  four  arcs,  70°,  80°. 
130°,  and  x.     Find  x  and  the  angles  of  the  quadrilateral  formed  by  the 
chords  of  these  arcs. 

16.  Find  the  angles  formed  by  the  diagonals  in  quadrilateral  of  No.  15. 

17.  Three  of  the  intercepted  arcs  of  a  circumscribed  quadrilateral  are 
68°,  98°,  114°.     Find  the  angles  of  the  quadrilateral.     If  the  chords  are 
drawn  connecting  (in  order)  the  four  points  of  contact,  find  the  angles  of 
this  inscribed  quadrilateral.    Also  find  the  angles  between  its  diagonals. 


108 


PLANE   GEOMETRY 


18.  If  the  angle  between  two  tangents  to  a  circle  is  40°,  find  the  other 
angles  of  the  triangle  formed  by  drawing  the  chord  joining  the  points  of 
contact. 

19.  The  circumference  of  a  circle  is  divided 
into  four  arcs,  three  of  which   are,    RS  =  62°, 
ST=  142°,  TU  =  98°.     Find  : 

(1)  Arc  UR; 

(2)  The  three  angles  at/2;  atS;  at  T\  at  [7; 

(3)  The  angles  A,  B,  C,  D  of  circumscribed 
quadrilateral ; 

(4)  The  angles  between  the  diagonals  R  T  and  SC7; 

(5)  The  angle  between  R U  and  ST  at  their  point  of  intersection 
(if  produced) ; 

(6)  The  angle  between  RS  and  TU  at  their  intersection; 

(7)  The  angle  between  AD  and  BC  at  their  intersection  ; 

(8)  The  angle  between  AB  and  DC  at  their  intersection ; 

(9)  The  angle  between  RS  and  DC  at  their  intersection; 
(10)  The  angle  between  AD  and  ST  at  their  intersection. 

20.  If  in  the  figure  of  No.  19,  Z  A  =  96° ;  Z  B  =  112° ;  and  Z  C  =  68°, 
find  the  angles  of  the  quadrilateral  RSTU.      [Denote  arc  RU  by  x. 
:.  in  A  ARU,  96°  +  |  x  +  £  a:  =  180°.     .'.  a;  =,  etc.] 

257.    It  is  evident  from  the  theorems  relating  to  the  measurement  of 
angles,  that : 

1    Equal  angles  are  measured  by  equal  arcs  (in  the  same  circle). 
2.  Equal  arcs  measure  equal  angles.  T 

21.  Prove  theorem  of  252  by  drawing  chord  parallel 
to  the  tangent. 

22.  The  opposite  angles  of   an  inscribed  quadri- 
lateral are  supplementary. 

Proof :  Z  A  +  Z  C  are  meas.  by  |  circumference. 

23.  If  two  chords  intersect  within  a  circle,  and  at 
right  angles,  the  sum  of  one  pair  of  opposite  arcs  equals 
the  sum  of  the  other  pair.  [Use  253.] 

24.  If  a  tangent  and  a  chord  are  parallel,  and  the  chords  of  the  two 
intercepted  arcs  be  drawn    they  wrill  make  equal  angles  with  the  tan- 
gent.    [Use  233 ;  252.] 

25.  The  line  bisecting  an  inscribed  angle  bisects  the  intercepted  arc. 


BOOK  II  109 

26.  The  line  joining  the  vertex  of  an  inscribed  angle  to  the  midpoint 
of  its  intercepted  arc  bisects  the  angle.  ...  p 

^*— ^r    — w 

27.  The  line  bisecting  the  angle  between  a  tangent 
and  a  chord  bisects  the  intercepted  arc. 

28.  State  and  prove  the  converse  of -No.  27. 

29.  The  angle  between   a  tangent  and  a  chord  is  half  the  angle 
between  the  radii  drawn  to  the  ends  of  the  chord. 

30.  If  a  triangle  be  inscribed  in  a  circle  and  a  tan- 
gent be  drawn  at  one  of  the  vertices,  the  angles  formed 
between   the   tangent   and   the   sides   will   equal   the 
other  two  angles  of  the  triangle. 

31.  By  the  figure  of  No.  30  prove  that  the  sum  of  the  three  angles  of 
a  triangle  equals  two  right  angles. 

32.  If  one  pair  of  opposite  sides  of  an  inscribed  quad- 
rilateral are  equal,  the  other  pair  are  parallel. 

Proof :  Draw  Js  BX,  CF;  arc  A  B  =  arc  CD  (?). 

'.  arc  ABC  =  arc  BCD  (Ax.  2). 

Hence  prove  rt.  &  ABX  and  CDY  equal. 

33.  If  any  pair  of  diameters  be  drawn,  the  lines  joining  their  extremi- 
ties (in  order)  will  form  a  rectangle.     [Use  251.] 

34.  If  two  circles  are  tangent  externally  and 
any  line  through  their  point  of  contact  intersects 
the  circumferences  at  B  and  C,  the  tangents  at  B 
and  C  are  parallel.     [Draw  common  tangent  at  A. 

Prove  :   Z  A  CT  =  Z  A BS.~\  v/ 

35.  Prove  the  same  theorem  if  the  circles  are  tangent  internally. 

36.  If  two  circles  are  tangent  externally  and  any  line  be  drawn  through 
their  point  of  contact  terminating  in  the  circumferences,  the  two  diam- 
eters drawn  to  the  extremities  will  be  parallel. 

37.  Prove  the  same  theorem  if  the  circles  are  tangent  internally. 

38.  If  two  circles  are  tangent  externally  and 
any  two  lines   be  drawn   through  their  point  of 
contact  intersecting  their  circumferences,  the  chords 
joining  these  points  of  intersection  will  be  parallel. 

[Draw  common  tangent  at  0.    Prove  :  Z  C  =  Z  /).] 

39.  Prove  the  same  theorem  if  the  circles  are  tangent  internally. 


110  PLANE   GEOMETRY 

40.  The  circle  described  on  one  of  the  equal  sides  of  an   isosceles 
triangle  as  a  diameter  bisects  the  base. 

Proof:  Draw  line  BM.     The  A  are  rt.  &  (?)  and  equal  (?). 

41.  If  the  circle,  described  on  a  side  of  a  triangle  as 
diameter,  bisects  another  side,  the  triangle  is  isosceles. 

42.  All  angles  that  are  inscribed  in  a  segment  greater 
than  a  semicircle  are  acute,  and  all  angles  inscribed  in  a 
segment  Jess  than  a  semicircle  are  obtuse. 

43.  An  inscribed  parallelogram  is  a  rectangle. 
Proof:  Arc  AB=  arc  CD  (?)  ;    arc  EC  =  arc  AD  (?). 
/.  arc  ABC '  =  arc  A  DC  (?);  that  is,  each  =  a  semicir- 

cumference.     Etc. 

44.  The  diagonals  of  an  inscribed  rectangle  pass  through  the  center 
and  are  diameters. 

45.  The  bisectors  of  all  the  angles  inscribed  in  the 
same  segment  pass  through  a  common  point. 

46.  The  tangents   at  the  vertices  of  an  inscribed 
rectangle  form  a  rhombus. 

[&  A  BF  and  HD  C  are  isosceles  (?)  and  equal  ?    Etc.] 

47.  If  a  parallelogram  be  circumscribed   about  a 
circle,  the  chords  joining  (in  order)  the  four  points  of 
contact  will  form  a  rectangle.     [Prove  BD  a  diameter.] 

48.  A  circumference  described  on  the  hypotenuse  of 
a  right  triangle  as  a  diameter  passes  through  the  vertices 
of  all  the  right  triangles  having  the  same  hypotenuse. 

49.  If  from  one  end  of  a  diameter  a  chord  be  drawn,  a  perpendicular  to 
it  drawn  from  the  other  end  of  the  diameter  will  inter- 
sect the  first  chord  on  the  circumference.     [Use  148.] 

50.  If  two  circles  intersect  and  a  diameter  be  drawn 
in  each  circle  through  one  of  the  points  of  intersection, 
the  line  joining  the  ends  of  these  diameters  will  pass 

through  the  other  point  of  intersection.    [Draw  chord  AB.    Use  251;  43.] 

51.  If  A  BCD  is  an  inscribed  quadrilateral,  AB  and  DC  produced  to 
meet  at  E,  AD  and  BC  produced  to  meet  at  F,  the  bisectors  of  angles 
E  and  F  are  perpendicular. 

[Show  that  the  circumference  is  divided  into  eight  arcs  which  are 
equal  in  pairs ;  by  correctly  adding  these,  show  that  four  arcs  are  equal 
to  the  others,  and  hence  equal  to  a  semicircumference.  The  angle  be- 
tween the  bisectors  is  measured  by  half  the  sum  of  these  four,  and  hence 
it  is  a  right  angle.] 


BOOK    II 


111 


52.  If  a  tangent  be  drawn  at  one  end  of  a  chord,  the 
midpoint  of  the  intercepted  arc  will  be  equally  distant 
from  the  chord  and  tangent. 

[Draw  chord  A  M  and  prove  the  rt.  &  = .] 

53.  If  two  circles  are  tangent  at  A  and  a  common 
tangent  touches  them  at  B  and  C,  the  angle  BAG  is  a 
right  angle.     [Draw  tangent  at  A.     Use  219;  148.] 

54.  A   circle  described  on  the  radius  of  another 
circle  as  diameter  bisects  all  chords  of  the  larger  circle 
drawn  from  their  point  of  contact.     To  Prove:    AB  is 
bisected  at  C.     Draw  chord  OC.     (Use  251 ;  213.) 

55.  If  the  opposite  angles  of  a  quadrilateral  are 
supplementary,  a  circle  can  be  drawn  circumscribing  it. 

To  Prove :    A  O  can  be  drawn  through  A,  B,  C,  P. 

Proof:  A  O  can  be  drawn  through  A,  B,  C  (?).     It 
is  required  to   prove  that  it  will  contain  point   P. 
Z  P  +  Z  B  are  supp.  (?).   .*.  must  be  meas.  by  half  the 
entire  circumf.     Z  B  is  meas.  by  £  arc  ADC  (V).     Hence 
Z  P  is  meas.  by  \  arc  ABC.     If  Z  P  is  within  or  without 
the  circumf.  it  is  not  meas.  by  \  arc  ABC.     (Why  not?) 

56.  The  circle  described  on  the  side  of  a  square,  or  of 
a  rhombus,  as  a  diameter  passes  through  the   point  of 
intersection  of  the  diagonals.     [Use  141;  148.] 

57.  The  line   joining  the   vertex   of  the  right   angle 
of  a  right  triangle  to  the  point  of  intersection  of  the 
diagonals  of  the  square  constructed  upon  the  hypotenuse 
as  a  side,  bisects  the  right  angle  of  the  triangle. 

Proof :    Describe  a  O  upon  the  hypotenuse  as  diameter 
and  use  148  ;  209  ;  249. 

58.  If  two  secants,  PAB  and  PCD,  meet  a  circle  at  A,  B,  and  C, 
D  respectively,  the  triangles  PBC  and  PAD  are  mutually  equiangular. 


59.  If  PAB  is  a  secant  and  PM  is  a  tangent  to 

a  circle  from  P,  the  triangles  PAM  &udPBM  are    B 
mutually  equiangular. 

60.  If  two  circles  intersect  and  a  line  be  drawn 
through  each  point  of  intersection  terminating  in 
the  circumferences,  the  chords  join  ing  these  extrem- 
ities will  be  parallel.     [Draw  RS.     Z  A  is  supp. 
of  Z  RSC  (?).     Finally  use  104.] 


112 


PLANE   GEOMETRY 


61.  If  two  equal  chords  intersect  within  a  circle, 

(1)  One  pair  of  intercepted  arcs  are  equal. 

(2)  Corresponding  parts  of  the  chords  are  equal.  c 

(3)  The   lines  joining  their  extremities    (in  order) 
form  an  isosceles  trapezoid. 

(4)  The  radius  drawn  to  their  intersection  bisects  their  angle. 

62.  If  a  secant  intersects  a  circumference  at  D  and  E, 
PC  is  a  parallel  chord,  and  PR  a  tangent  at  P  meeting 
secant  at  R,  the  triangles  PCD  and  PRD  are  mutually 
equiangular.  [Z  R  and  Z  CDP  are  measured  by  £  arc  RC. 
(Explain.)     Etc.] 

63.  If  a  circle  be  described  upon  one  leg  of  a  right 
triangle  as  diameter  and  a  tangent  be  drawn  at  the 
point  of   its  intersection  with  the   hypotenuse,  this 
tangent  will  bisect  the  other  leg. 

[Draw  OP  and  OD.  CD  is  tangent  (?).  OD 
bisects  arc  PC  (?)  (220).  Z  COD  =  Z  A  (?)  (257). 
/.  OD  is  ||  to  AB  (V).  Etc.] 

64.  If  an  equilateral  triangle  ABC  is  inscribed  in  a 
circle  and  P  is  any  point  of  arc  AC,  AP  +  PC  =  BP. 
[Take  PN  =  PA  ;    draw  AN.     &ANP  is  equilateral. 
(Explain.)     AANB  =  A  A  PC  (?).     Etc.] 

65.  If  two  circles  are  tangent  internally  at  C,  and  a 
chord  AB  of  the  larger  circle  is  tangent  to  the  less 
circle  at  M,  the  line  CM  bisects  the  angle  A  CB. 

[Draw  tangent  CX  and  chord  RS.  Z  KSC  = 
^BCX  =  ZA.  .-.  AB  is  ||  to  RS.  (Explain.) 
Then  use  233.  Etc.] 

66.  If  two  circles  intersect  at  A   and   C  and 
lines  be  drawn  from  any  point  P,  in  one  circum- 
ference, through  A   and  C  terminating  in  the  other 
at  points  B  and    D,   chord  BD    will  be   of  constant  p 
length  for  all  positions  of  point  P. 

[Draw  BC.     Prove  Z  BCD,  the  e*xt.  Z  of 
=  a  constant.     Etc.] 

67.  The  perpendiculars  from  the  vertices  of 
a  triangle  to  the  opposite  sides  are  the  bisectors 
of  the  angles  of  the  triangle  formed  by  joining 
the  feet  of  these  perpendiculars. 

To  Prove:  BS  bisects  ^RST,  etc. 


B 


HOOK    II  11:1 

Proof:  If  a  circle  be  described  on  AO  as  diam.,  it  will  pass  through  T 
and  ^  (?)  (148).  If  a  circle  be  described  on  OC  as  diam.,  it  will  pass 
through  R  and  5  (?).  .'.  ZBAR  =  ^BST(!)  ;  and  /.BCT  =  ^BSR  (?). 
But  Z BAR  =  Z  BCT.  (Each  is  the  comp.  of  Z  ABC.  . .  Etc.) 

68.  If  ABC  is  a  triangle  inscribed  in  a  circle,  BD  is 
the  bisector  of  angle  ABC,  meeting  AC  at  O  and  the 
circumference  at  D,  the  triangles  A  OB  and  COD  are 
mutually  equiangular.     Also  triangles  BOC  and  A  OD. 
Also  triangles  BOC  and  ABD.    Also  triangles  A  OD  and 
ABD.     Also  triangles  £CZ>  and  COD. 

69.  If  two  circles  intersect  at  A  and  5,  and  from  P,  any  point  on  one 
of  them,  lines  A  P  and  BP  be  drawn  cutting  the  other  circle  again  at  C 
and  D  respectively,  CD  will  be  parallel  to  the  tangent  at  P. 

70.  If  two  circles  intersect  at  A  and  B,  and  through  B  a  line  be  drawn 
meeting  the  circles  at  R  and  S  respectively,  the  angle  RAS  will  be 
constant  for  all  positions  of  the  line  RS. 

[Prove  Z  R  +  Z  S  is  constant.     .'.  Z.  RAS  is  also  constant.] 

71.  Two  circles  intersect  at  A  and  through  A  any  secant  is  drawn 
meeting  the  circles  again  at  M  and  N.    Prove  that  the  tangents  at  M  and 
N  meet  at  an  angle  which  remains  constant  for  all  positions  of  the 
secant. 

[Prove  the  angle  between  these  tangents  equal  to  the  angle  between 
the  tangents  to  the  circles,  at  A .] 

72.  Three  unequal  circles  are  each  externally  tangent  to  the  other 
two.     Prove  that  the  three  tangents  drawrn  at  the  points  of  contact  of 
these  circles  meet  in  a  common  point. 

73.  Two  equal  circles  intersect  at  A   and  B,  and  through  A   any 
straight  line  MAN  is  drawn,  meeting  the  circumferences  at  M  and  N 
respectively.     Prove  chord  BM  =  chord  BN. 

74.  If  the  midpoint  of  the  arc  subtended  by  any  chord  be  joined  to 
the  extremities  of  any  other  chord, 

(1)  The  triangles  formed  will  be  mutually  equiangular. 

(2)  The  opposite  angles  of  the  quadrilateral  thus  formed  will  be  sup- 
plementary. 

75.  Two  circles  meet  at  A   and  B  and  a  tangent  to  each  circle  is 
drawn  at  A,  meeting  the  circumferences  at  R  and  S  respectively;  prove 
that  the  triangles  ABR  and  ABS  are  mutually  equiangular. 

BOBBINS'  PLANE  GEOM.  — 8 


114 


PLANE   GEOMETRY 


76.  What  is  the  locus  of  points  at  a  given  distance  from  a  given 
point?     Prove.     (Review  179  and  180  now.) 

77.  What  is  the  locus  of  the  midpoints  of  all  the  radii  of  a  given 
circle  ?    Prove. 

78.  What  is  the  locus  of  the  midpoints  of  a  series  of  parallel  chords 
in  a  circle  ?     Prove. 

79.  What  is  the  locus  of  the  midpoints  of  all  chords  of  the  same 
length  in  a  given  circle  ?    Prove. 

80.  What  is  the  locus  of  all  points  from  which  two  equal  tangents 
can  be  drawn  to  two  circles  which  are  tangent  to  each  other? 

81.  What  is  the  locus  of  all  the  points  at  a  given  distance  from  a  given 
circumference?    Discuss  if  the  distance  is  >  radius.     If  it  is  less. 

82.  What  is  the  locus  of  the  vertices  of  the  right  angles  of  all  the 
right  triangles  that  can  be  constructed  on  a  given  hypotenuse?    Prove. 

83.  What  is  the  locus  of  the  vertices  of  all  the  triangles  which  have  a 
given  acute  angle  (at  that  vertex)  and  have  a  given  base?     Prove. 

84.  A  line  of  given  length  moves  so  that  its  ends  are  in  two  perpen- 
dicular lines ;  what  is  the  locus  of  its  midpoint  ?     Prove. 

[Suppose  AB  represents  one  of  the  positions  of 
the  moving  line.  Draw  OP  to  its  midpoint.  In  all 
the  positions  of  AB,  OP  =  \  AB  =  a  constant  (148). 

.'.  P  is  always  a  fixed  distance  from  0.     Etc.] 

85.  What  is  the  locus  of  the  midpoints  of  all  the 

chords  that  can  be  drawn  through  a  fixed  point  on  a  given  circumfer- 
ence ?    Prove. 

[Suppose  AB  represents  one  of  the  chords  from  B 
in  circle  0,  with  radius  OB ;  and  P  the  midpoint  of 
AB.  Draw  OP.  Z  P  is  a  rt.Z  (?).  That  is,  where  - 
ever  the  chord  may  be  drawn,  /.  P  is  a  rt.  Z. 

.'.  locus  of  P  is,  etc.] 

86.  A  definite  line  which  is  always  parallel  to  a  given  line  moves  so 
that  one  of  its  extremities  is  on  a  given  circumference ;  find  the  locus  of 
the  other  extremity. 

[Suppose  CP  represents  one  position  of 
the  moving  line  CP.  Draw  OQ  =  and  II 
to  CP  from  center  0.  Join  OC  and  PQ. 
Wherever  CP  is,  this  figure  is  a  O  (?).  Its 
sides  are  of  constant  length  (?).  That  is,  P 
is  always  a  fixed  distance  from  Q,  etc.]  A- 


O 


BOOK   11  115 

CONSTRUCTIONS 

258.  Heretofore  the  figures  we  have  used  have  been  assumed.     We 
have  supposed  such  auxiliary  lines  to  be  drawn  as  conditions  required. 
No  methods  have  been  given  for  drawing  any  lines,  and  only  our  three 
postulates  have  been  assumed  regarding  such  construction.    But  the  lines 
that  have  been  used  were  drawn  as  aids  toward  establishing  truths,  and 
precise  drawings  have  not  been  essential.     The  following  simple  methods 
for  constructing  lines  are  given  that  mathematical  precision  may  be  em- 
ployed if  necessary  in  drawing  diagrams  of  a  more  complex  nature.    The 
pupil  should  be  very  familiar  with  the  use  of  the  ruler  and  compasses. 

259.  A   geometrical   construction   is   a   diagram  made  of 
points  and  lines. 

260.  A  geometrical  problem  is  the  statement  of  a  required 
construction.     Thus:   "it  is  required  to  bisect  a  line"  is  a 
problem.     A  problem  is  sometimes  defined  as  "a  question  to 
be  solved  "  and  includes  other  varieties  besides  those  involved 
in  geometry. 

261.  The  word  proposition  is  used  to  include  both  theorem 
and  problem. 

262.  The  complete  solution  of  a  problem  consists  of  five 
parts  : 

I.    The  Given  data  are  to  be  described. 
II.    The  Required  construction  is  to  be  stated. 

III.  The  Construction  is  to  be  outlined. 

This  part  usually  contains  the  verb  only  in  the  imperative. 
No  reasons  are  necessary  because  no  statements  are  made. 
The  only  limitation  in  this  part  of  the  process  is,  that  every 
construction  demanded  shall  have  been  shown  possible  by 
previous  constructions  or  postulates.  (See  32;  33;  199.) 

IV.  The  Statement  that  the  required    construction   has 
been  completed. 

V.    The  Proof  of  this  declaration. 

Sometimes  a  discussion  of  ambiguous  or  impossible  in- 
stances will  be  necessary. 


116 


PLANE   GEOMETRY 


IM 


263.  NOTES.   (1)  A  straight  line    is  determined  by  two 
points. 

(2)  A  circle  is  determined  by  three  points. 

(3)  A  circle  is  determined  by  its  center  and  its  radius. 
Whenever  a  circumference,  or  even  an  arc,  is  to  be  drawn,  it 
is  essential  that  the  center  and  the  radius  be  mentioned. 

(4)  "Q.E.F."  =  Quod   erat  faciendum  =  "  which  was  to  be 
done."     These  letters  are  annexed  to  the  statement  that  the 
construction  which  was  required  has  been  accomplished. 

264.  PROBLEM.    It  is  required  to  bisect  a  given  line. 
Given  :   The  definite  line  AB. 

\  :  ..-• 

Required  :  To  bisect  AB. 

Construction  :  Using  A  and  B 
as  centers  and  one  radius,  suffi- 
ciently long  to  make  the  cir- 
cumferences intersect,  describe 
two  arcs  meeting  at  R  and  r. 
Draw  RT  meeting  AB  at  M. 

Statement:  Point  M  bisects 
AB.  Q.E.F. 

Proof :  R  is  equally  distant  from  A  and  B  (?)  (201). 

T  is  equally  distant  from  A  and  B  (?). 

Hence,  RT  is  the  J.  bisector  of  AB  (?)  (70). 

That  is,  M  bisects  AB.  Q.E.D. 

265.  PROBLEM.    To  bisect  a  given  arc. 
Given  :  Arc  AB  whose  center 

is  O. 

Required  :  To  bisect  arc  AB. 

Construction:  Draw  chord 
AB ;  using  A  and  B  as  centers 
and  any  sufficient  radius,  de- 
scribe arcs  meeting  at  C.  Draw 
OC  cutting  arc  AB  at  M. 


O-- 


BOOK   TI 


117 


Statement:  The  point  M  bisects  arc  AB.  Q.E.F. 

Proof :   o  and  C  are  each  equally  distant  from  A  and  B  (201). 
.*.  OC  is  the  _L  bisector  of  chord  AB  (?)  (70). 
.-.  M  bisects  arc  AB  (?)  (213).  Q.E.D. 

266.  PROBLEM.    To  bisect  a  given  angle. 
Given:    Z  LON. 

Required  :    To  bisect  Z  LON. 

Construction :  Using  O  as  a 
center  and  any  radius,  draw  arc 
AB,  cutting  LO  at  A  and  NO  at 
B.  Draw  chord  AB.  Using  A 
and  B  as  centers  and  any  suf- 
ficient radius,  draw  two  arcs 
intersecting  at  S.  Draw  OS  meeting  arc  AB  at  M. 

Statement:    OS  bisects  Z  LON.  Q.E.F. 

Proof :    O  and  S  are  each  equally  distant  from  A  and  B  (?). 

.-.  OS  is  the  _L  bisector  of  chord  AB  (?). 

.-.  M  bisects  arc  AB  (?)  (213).  .-.  Z  AOM  =  Z  BOM  (?)  (207). 

That  is,  OS  bisects  Z  LON.  Q.E.D. 

267.  PROBLEM.    At  a  fixed  point  in  a  straight  line  to  erect  a  perpen- 
dicular to  that  line. 

Given  :   Line  A  B  and  point 
P  within  it. 

Required:   To  erect  a  line 
_L  to  AB  at  P. 

Construction:  Using  Pas  A— 
a  center  and  any  radius, 

draw  arcs  meeting  AB  at  C  and  D.  Using  C  and  D  as  centers 
and  a  radius  longer  than  before,  draw  arcs  meeting  at  S. 
Draw  PS. 

Statement :    PS  is  _L  to  AB  at  P.  Q.E.F. 

Proof :    PS  is  the  _L  bisector  of  CD  (?)  (70).  Q.E.D. 


•D 


•B 


118 


PLANE   GEOMETRY 


Another  Construction :  Using  any  point  O,  without  AB,  as 
center,  and  OP  as  radius, 
describe  a  circumference, 
cutting  AB  at  P  and  E. 
Draw  diameter  EOS. 
Join  SP. 

Statement :  -SP  is  JL  to 
AB  at  P.  Q.E.F. 

Proof :    Segment   SPE  A  EX.  .xp 

is  a  semicircle  (?)  (204). 

.-.  Z  SPE  is  a  rt.  Z  (?)  (251).     .-.  SP  is  J.  to  AB  (?)  (17). 

268.  PROBLEM.   Through  a  point  without  a  line  to  draw  a  perpen- 
dicular to  that  line.  P 

Given :    Line     AB     and 
point  P  without  it.  si/ 

Required:    (?).  /f\ 

Construction :    Using    P 
as  a  center  and  any  suffi-     ^        --M 

cient   radius,  describe    an  " 

arc  intersecting  AB  at  M  and  JV.  Using  Jf  and  N  as  centers 
and  any  sufficient  radius,  describe  arcs  intersecting  each  other 
at  C.  Draw  PC. 

Statement:    PC  is  J_  to  AB  from  P.  Q.E.F. 

Proof :    PC  is  the  J_  bisector  of  MN  (?)  (70).  Q.E.D. 

269.  PROBLEM.    At  a  given  point  in  a  given  line  to  construct  an  angle 
which  shall  be  equal  to  a  given  angle. 

Given:    Z.AOB;    point   P   in 
line  CD. 

Required :    To  construct  at  P  o 
an  Z  =  Z  AOB. 

Construction:    Using    O   as  a 
center  with  any  radius,  describe 
an  arc  cutting  OA  at  E  arid  OB    c 


BOOK  II  119 

at  F.  Draw  chord  EF.  Using  P  as  a  center  and  OE  as  a 
radius,  describe  an  arc  cutting  CD  at  R.  Using  B  as  a  center 
and  chord  EF  as  a  radius,  describe  an  arc  cutting  the  former 
arc  at  X.  Draw  PX  and  chord  RX. 

Statement :    The  Z  XPD  =  Z  AOB.  Q.E.F. 

Proof:    Chord EF=  chord  EX  (?)  (201). 

.-.  arc  EF=  arc  RX  (?)  (209).  .-.  Z  XPR  =  Z  O  (?)  (207). 
That  is,  Z  XPD  =  Z  ^L(XB.  Q.E.D. 

270.  PROBLEM.  To  draw  a  line  through  a  given  point  parallel  to  a 
given  line. 

Given :    Point  P  and  line  AB.     \ 

Required :    To  draw  through     p\  X 

P,  a  line  II  to  AB. 


Construction:  Draw  any  line  \ 

PN  through  P  meeting  AB  at  N.          ~"N~~ 

On  this  line,  at  P,  construct  Z  NPX=Z.  ANP.     (By  269.) 
Statement:   PX  is  II  to  AB.  Q.E.F. 

Proof:    ZNPX=/.ANP   (?)    (Const.). 
/.  PX  is  II  to  AB  (?)  (101).  Q.E.D. 

271.   PROBLEM.   To  divide  a  line  into  any  number  of  equal  parts. 

Given :    Definite  line  AB.     A         O        N        M         L         B 

'*••«.     / 
Required:    To     divide     it       "'"^. 

into  five  equal  parts. 

Construction:  Draw 
through  A  any  other  line  AX. 
On  this  take  any  length  AC 
as  a  unit,  and  mark  off  on  AX 
five  of  these  units,  AC,  CD,  DE,  EF,  FG.  Draw  GB. 

Through  F,  E,  D,   C,  draw  II  to  GB,  lines  FL,  EM,  DN,  CO. 

Statement :  Then,  AO  =  ON  =  NM=  ML  =  LB.  Q.E.F. 

Proof  :    AC  =  CD  =  DE  =  EF  =  FG  (Const.). 

.'.A0=  ON=NM=ML  =  LB  (?)  (147).  Q.E.D. 


120  PLANE   GEOMETRY 

272.   PROBLEM.    To  draw  a  tangent  to  a  given  circle  through  a  given 
point: 

I.  If  the  point  is  on  the  circumference. 

II.  .If  the  point  is  without  the  circle. 

I.  Given  :  O  O ;  P,  a  point      A" 
011  the  circumference. 

Required:  To  draw  a 
tangent  through  P. 

Construction:  Draw  the 
radius  OP.  Draw  line  AB  J_ 
to  OP  at  P  (by  267). 

Statement :  AB  is  tangent 
to  O  O  at  P.  Q.E.F. 

Proof :    AB  is  _1_  to  PO  at  P  (Const.). 

.-.  AB  is  a  tangent  (?)  (215).  Q.E.D. 

II.  Given  :  O  O ;  P,  a  point 
without  it. 

Required :   To  draw  a  tan-       . 
gent  through  P. 


Construction:   Draw   PO;       \ 
bisect  it  at  M  (by  264). 

Using  M  as  a  center  and 
PM  as  a  radius,  describe  a 
circumference  intersecting  O  O  at  A  and  B. 

Draw  PA,  PJ5,  OA,  OB. 

Statement :    PA  and  PB  are  tangents  through  p.        Q.E.F. 
Proof :  O  M  passes  through  O  (P3f  =  MO  by  const.). 

.-.  Z  P^LO  is  a  rt.  Z  (?)  (251).     /.  PA  is  tangent  (?)  (215). 
Similarly  PB  is  a  tangent.  Q.E.D. 

Ex.  1 .    Show  by  two  distinct  methods  how  to  bisect  a  line. 

Ex.  2.    Show  how  to  construct  the  problem  of  270  by  use  of  268. 


BOOK  II 


121 


273.  PROBLEM.  To  circumscribe  a  circle  about  a  given  triangle. 
Given:  (?).     Required:   (?). 

(See  227.) 

Construction  :  Bisect  AB,  BC, 
AC.  Erect  Js  at  these  midpoints, 
meeting  at  O. 

Using  O  as  a  center  and  OA 
as  radius,  draw  a  circle. 

Statement:  This  O  will  pass 
through  vertices  -4,  fi,  and  C. 

Q.E.F. 

Proof:    [Use  85.] 

274.  PROBLEM    To  inscribe  a  circle  in  a  given  triangle. 


Given:  (?).  Required:  (?).  Construction:  Draw  the 
three  bisectors  of  the  A  of  A  ABC,  meeting  at  O  (by  266). 
Draw  Js  from  O  to  the  three  sides. 

Using  O  as  a  center  and  one  of  these  Js  as  a  radius,  draw  a  O. 

Statement :  THis  O  will  be  tangent  to  the  three  sides  of 
A  ABC.  Q.E.F. 

Proof  :  The  bisectors  of  these  angles  meet  in  a  point  and 
the  Js  OL,  OM,  ON  are  equal  (?)  (84). 

/.  the  circumference   passes  through  L,  M,  N  (?)  (192). 

Therefore  the  three  sides  are  tangent  to  the  O  (?)  (215). 

That  is,  the  O  O  is  inscribed  in  A  ABC  (?)  (234).        Q.E.D. 


122 


PLANE   GEOMETRY 


275.   PROBLEM.  To  construct  a  parallelogram  if  two  sides  and  the 


w 


s y 


Q.E.F. 


included  angle  are  given. 

Given  :  The  sides  a  and  b  and 
their  included  angle,  x. 

Required :     To    construct    a 
O  containing  these  parts. 

Construction :  Take  a  straight 
line  PQ  z=  a. 

At    P    construct   Z  P  =  Z  x.     g 

On    PW,   the   side    of    this  Z,     ^ 

take  PR  =  b. 

At  B  draw  BY  \\  to  PQ;  and  at  Q  draw  QZ  ||  to  PW. 

Denote  the  intersection  of  these  lines  by  8. 

Statement :    PQSB  is  the  required  parallelogram. 

Proof  :    First,  it  is  a  parallelogram  (?). 

Second,  it  is  the  required  parallelogram.     (Because  it  con- 
tains the  given  parts.)  Q.E.D. 

276.   PROBLEM.  To  construct  a  segment  of  a  circle  upon  a  given 
line,  as  chord,  which  shall  contain  angles  equal  to  a  given  angle. 

Given :  Line  AB  and 
Zir'. 

Required :  To  con- 
struct a  segment  upon  AB 
whose  inscribed  angles 
shall  =  Z  K*. 

Construction:  Con- 
struct at  A,  /-BAC=Z.K'. 

Bisect  AB  at  M. 

At  M  erect  OM  _L  to  AB. 

At  A  erect  OA  _L  to 
AC,  meeting  OM  at  O. 

Using  O  as  a  center  and  OA  as  radius,  describe  O  O. 

Statement :    The  A  inscribed  in  AKB  =  Z  Kr.  Q.E.F. 


BOOK    II  123 

Proof:  Tin?  circumference  passes  through  B  (?)  (67). 
/.  AB  is  u  chord  (?).  AC  is  tangent  to  the  O  (?)  (215). 
/.  /iBACis  -measured  by  half  the  arc  AB  (?)  (252). 

Any  inscribed  angle  AKB  is  measured  by  half  the  arc  ^#(?). 

Therefore,  any  angle  AKB  =  Z  BAG  (?)  (257,  2). 

Consequently,  any  inscribed  angle  AKB  =  Z  K1  (Ax.  1). 

Q.E.D. 

[If  the  pupil  will  draw  chords  AK  and  BK,  he  will  under- 
stand the  proposition.  These  were  purposely  omitted.] 

277.  PROBLEM.  To  construct  the  third  angle  of  a  triangle  if  two 
angles  are  known. 

Given  :    A  A  and  .B,  two  A 

of  Ji  A. 

Required :  To  construct  the 
third. 

Construction  :  At  point  O  in 
a  line  ES  construct  Z  a  =  Z  A.    R~~  o" 

At  point  O  in  OT  construct  Z  b  =  Z  B. 
Statement :   The  Z  VOE  =  the  third  Z  of  the  A.         Q.E.F. 
Proof  :    Za  +  Z£  +  Z  VOE  =  2  rt,  A  (?)  (46). 
/.A  +  /.B  +  the  unknown  Z  =  2  rt.  ^  (?)  (110). 
.-.  Z  a  +  Z  b  +Z  F(XR  =  Z  A  +  Z  £  +  the  unknown  Z  (?). 

But  Z  a  +  Z5 =  Z  A  +  Z  B  (Const,  and  Ax.  2). 

.'.  Z  FO.B       =  the  unknown  Z  (Ax.  2). 
That  is,  ZFOtf  =  the  third  Z  of  the  A.  Q.E.D. 


Ex.  1.   To  circumscribe  a  triangle  about  a  given  circle. 

Ex.  2.  To  construct  the  problem  of  276  if  the  given  angle  is  a  right 
angle ;  if  it  is  an  obtuse  angle. 

Ex.  3.    To  construct  the  problem  of  273  if  the  given  triangle  is  obtuse. 

Ex.4.   Is  the  problem  of  277  ever  impossible  ?     Explain. 

Ex.  5.  In  the  figure  of  274,  if  Z  A  =  40°  and  /.B  =  94°,  how  many 
degrees  are  there  in  each  of  the  six  acute  angles  at  0  ?  If  A  LMN 
is  constructed,  how  many  degrees  are  there  in  each  of  its  angles  ? 


124 


PLANE   GEOMETRY 


278.  PROBLEM.  To  construct  a  triangle  if  the  three  sides  are  known. 
Given :    Sides 

a,  6,  c  of  a  A.      . 

Required:  To  Ct 
construct  the  A. 

Construction  : 

Draw  RS  =  a. 

^  ff 

Using  R  as  a  R 

center  and  b  as  a  radius,  describe  an  arc ;  using  S  as  a  center 
and  c  as  a  radius,  describe  an  arc  intersecting  the  former  arc 
at  T.  Draw  ET  and  ST. 

Statement:    RST  is  the  required  A.  Q.E.F. 

Proof:    RST  is  a  A  (?)  (23). 

RST  is  the  required  A.    (It  contains  a,  5,  <?.)  Q.E.D. 

Discussion  :    Is  this  problem  ever  impossible  ?    When  ? 

279.  PROBLEM.  To  construct  a  triangle  if  two  sides  and  the  included 
angle  are  known.  g 

Given  :    The  sides  a  and  6,        *> 
and  their  included  Z  C  in  a  A. 

Required:     To    construct 
the  A. 

Construction  :  Draw  CB  —a. 
At  C  construct  Z  BCX  =  given    c 
Z  C.     On  ex  take  CM  =  b.     Join 

Statement:   (?).     Proof:    (?). 


NOTE.  The  student  has  probably  observed  that  in  constructions  cer- 
tain lines  and  angles  must  precede  others.  In  such  problems  as  266,  267, 
269,  272,  and  276,  the  order  of  the  successive  steps  is  exceedingly  impor- 
tant. Problems  are  not  so  numerous  in  geometry  as  theorems,  but  it  must 
be  apparent  that  problems  are  instructive,  fascinating,  and  profitable. 

Definition.  If  a  circle  is  described,  touching  one  side  of  a 
triangle  and  the  prolongations  of  the  other  sides,  it  is  called 
an  escribed  circle. 


BOOK    II 


125 


280.   PROBLEM.  To  construct  a  triangle  if  a  side  and  the  two  angles 
adjoining  it  are  known. 


\ 


Given:   (?).    Required:   ('0-  B< 

Construction :  Draw  BC  =  a.  At  B  construct  Z  CBX  —  Z  B1 ; 
at  C  construct  Z  BCY  =  Z  c'.  Denote  the  point  of  intersec- 
tion of  BX  and  CY  by  A. 

Statement:    (?).     Proof:    (?).     Discussion:    (?). 

281.  PROBLEM  To  construct  a  right  triangle  if  the  hypotenuse  and 
a  leg  are  known. 

Given:  Hypotenuse*?;  leg b.    A 

Required:    (?)! 

Construction:  Draw  an  in- 
definite line  CD  and  at  C  erect 
a  _L  =  b.  Using  A  as  a  center 
and  c  as  a  radius,  describe  an 
arc  cutting  CD  at  B.  Draw  AB. 

Statement:  (?).     Proof  :  (?). 

NOTE.  If  it  is  required  to  construct  a  right  triangle,  having  given  the 
hypotenuse  and  another  part,  it  is  often  advantageous  to  describe  a  semi- 
circle upon  the  given  hypotenuse  as  diameter.  Every  triangle  whose  base 
is  this  diameter  and  whose  vertex  is  on  this  semicircumference  is  a  right 
triangle  (251).  Hence  if  the  triangle  constructed  contains  the  other 
given  part,  it  is  the  required  triangle. 

Ex.  1.  To  construct  the  problem  of  281  by  use  of  the  semicircle. 

Ex.  2.  Discuss  the  constructions  of  279,  280,  and  281  fully. 

Ex.  3.  To  construct  a  triangle  and  its  three  escribed  circles. 

Ex.  4.  To  construct  an  isosceles  right  triangle  having  given  the  hypot- 
enuse. 

Ex.  5.  To  construct  an  isosceles  right  triangle  having  given  the  leg. 

Ex.  6.  To  construct  a  quadrilateral  having  given  the  four  sides  and 
an  angle. 


Discussion:  (?). 


126 


PLANE   GEOMETRY 


282.  PROBLEM.  To  construct  a  triangle  if  an  angle,  a  side  adjoin- 
ing it,  and  the  side  opposite  it  are  known ;  that  is,  if  two  sides  and 
an  angle  opposite  one  of  them  are  known. 

The  known  angle  may  be  obtuse,  right,  or  acute.   Consider : 
First,  If  "side  opposite"  >  "side  adjoining." 
Second,  If  "side  opposite "=  "side  adjoining." 
Third,  If  "side  opposite  "  <  "side  adjoining." 
Construction  for  all  of  these :    Draw  an  indefinite  line,  ITX, 
and  at  one  extremity  construct  an  Z  =Z  K  ;   take  on  the  side 
of  this  /-  a  distance  from  the  vertex  =  "side   adjoining." 
Using  the  end  of  this  side  as  a  center  and  the  "side  opposite  " 
as  a  radius,  describe  an  arc  intersecting  KX.     Draw  radius 
to  the  intersection  just  found. 

If  the  known  angle  is  obtuse  or  right 

Given :  Z  K,  s.a,  and  s.o.  of  a  A.  v 

Construction :  As  above. 

Discussion:  Case  I.  s.o.  >s.#. 
The  A  is  always  possible. 

Case  II.  s.o.  =s.  a.  The  A  is 
never  possible  (55  and  112). 

Case  III.  s.  o.  <  s.  a. 
The  A  is  never  possible 
(?)  (122). 

If  the  known  angle  is 
acute. 

Case      I.  s.  o.  >s.  a. 
The  A  is  always  possible. 

Case    II.    s.o.  =  8. a. 
isosceles  A. 

Case  III.   s.  o.  <  s.  a. 

(1)  If  s.o.  <  the  _L  from 
A  to  jBTX,  the  A  is  never  pos- 
sible. 


BOOK   II 


127 


(2)  If  s.o.  =  the  _L  from 
A  to  ZX,  A  is  a  rt.A  (216). 

(3)  If  s.o.  >  the  -L  from 
A  to  KX,  there  are  two  A. 

In  this  instance  the  arc 
described,  using  A  as  cen- 
ter and  "  s.o"  as  radius,  in- 
tersects KX  twice,  at  B 
and  Bf. 

Hence,  A  AKB  and  A  AKB1 
both  contain  the  three  given 
parts. 

CONCERNING  ORIGINAL   CONSTRUCTIONS 
ANALYSIS 

Many  constructions  are  so  simple  that  their  correct  solution  will 
readily  occur  to  the  pupil.  Sometimes,  as  in  the  case  of  complicated  con- 
structions, one  requires  the  ability  to  put  the  given  parts  together,  one 
by  one. 

The  following  outline  may  be  found  helpful  if  employed  intelligently. 
I.  Suppose  the  construction  made,  —  that  is,  suppose  the  figure  drawn. 

II.  Study  this  figure  in  search  of  truths  by  which  the  order  of  the 
lines  that  have  been  drawn  can  be  determined.  This  is  essential. 

III.  One  or  more  auxiliary  lines  may  be  necessary. 

IV.  Finally,  construct  the  figure  and  prove  it  correct. 

EXERCISE.  Given  the  base  of  a  triangle,  an  adjacent  acute  angle, 
and  the  difference  of  the  other  sides,  to  construct  the  triangle. 

Given:  Base^ljB;  /.A'\  differenced. 

Required  :  To  construct  the  A.  d 

[Analysis  :  Suppose  A  AE  C  is  the  required  A.      ( 
It  is  evident  that  if  CD  =  CB,  they  may  be  sides 
of  an  isos.  A  and  AD  =  d.  This  isosceles  A  may 
be  constructed.] 


Construction:  At  A  on  AE  construct 
ZBA\  =  ^A'  and  take  on  AX,  AD  =  d. 
Join  DB.  At  Af,  midpoint  of  DB,  draw  M  Y 
±  to  DB  meeting  AX  at  C.  Draw  CB. 

Statement  :  (?)  .  Proof  :  (?)  .  Discussion  :  (?)  . 


rx 


128  PLANE   GEOMETRY 

ORIGINAL    CONSTRUCTIONS 

1.  To  construct  a  right  angle. 

2.  To  construct  an  angle  containing  45°. 

3.  To  construct  the  complement  of  a  given  angle ;  the  supplement. 

4.  To  construct  an  angle  of  60°.     [See  115.] 

5.  To  construct  an  angle  of  30°  ;  of  15° ;  of  120°. 

6.  To  construct  an  angle  of  150°;  of  135° ;  of  75°  ;  of  165°. 

7.  To  find  the  center  of  a  given  circle.     [See  214.] 

8.  To  construct  a  tangent  to  a  given  circle,  parallel  to  a  given  line. 
[Draw  a  radius  _L  to  the  given  line.] 

9.  To  construct  a  tangent  to  a  given  circle,  perpendicular  to  a  given 
line. 

10.  To  construct  the  other  acute  angle  of  a  right  triangle  if  one  is 
known. 

11.  To  draw  through  a  given  point  without  a  given  line,  another  line 
which  shall  make  a  given  angle  with  the  line. 

[Draw  a  ||  to  the  given  line  through  the  given  point.] 

12.  To  trisect  a  right  angle. 

13.  To  find  a  point  in  one  side  of  a  triangle  equally  distant  from  the 
other  sides.     [Use  266.] 

14.  To  construct  a  chord  of  a  circle  if  its  midpoint  is  known. 
[Draw  a  radius  through  this  point  and  use  267.] 

15.  To  construct  the  shortest  chord  that  can  be  drawn  through   a 
given  point  within  a  circle. 

Proof:  Draw  any  other  chord  through  the  point,  etc. 

16.  To  construct  through  a  given  point  within  a 
circle  two  chords  each  equal  to  a  given  chord. 

17.  To  construct  in  a  given  circle  a  chord  equal  to 
a  second  chord  and  parallel  to  a  third. 

18.  To  construct  through  a  given  point  a  line 
which  shall  make  equal  angles  with  the  sides  of  a 
given  angle.     [Use  266  ;  268.] 

19.  To  construct  from  a  given  point  in  a  given 
circumference  a  chord  which  shall  be  at  a  given 
distance  from  the  center. 

[How  many  can  be  drawn  from  this  point?] 


BOOK  ir 


129 


1.   To  construct  an  isosceles  triangle,  having  given: 

20.  The  base  and  one  of  the  equal  sides. 

21.  The  base  and  one  of  the  equal  angles. 

22.  One  of  the  equal  sides  and  the  vertex-angle. 

23.  One  of  the  equal  sides  and  one  of  the  equal  angles. 
24    The  base  and  altitude  upon  it. 

25.  The  base  and  the  radius  of  the  inscribed  circle. 
[Bisect  the  base;  erect  a  JL  =  radius;  describe  0,  etc.] 

26.  The  base  and  the  radius  of  the  circumscribed  circle. 
[First,  describe  O  with  given  radius  and  any  center.] 

27.  The  altitude  and  the  vertex-angle. 

[Draw  an  indefinite  line  and  erect  a  _L  equal  the  given  altitude.     Bisect 
the  given  Z ;  at  the  end  of  the  altitude  construct  Z  =  £  given  Z,  etc.] 

28.  The  base  and  the  vertex-angle. 

[Find  the   supplement  of  given  Z ;  bisect  this ;  at  each  end  of  base 
construct  an  Z  =  this  half ;  etc.] 

29.  The  perimeter  and  the  altitude. 
Given:     Perimeter  =  AB;       alt.  =  h. 

Required:   (?).  Construction:   Bisect  AB, 
erect  at  M  ±  =  A;  draw  AP  and  BP. 
Bisect  these ;  erect  _b  SC  and  RE ;  etc. 


II.  To  construct  a  right  triangle,  having  given : 

30.  The  two  legs. 

31.  One  leg  and  the  adjoining  acute  angle. 

32.  One  leg  and  the  opposite  acute  angle. 

33.  The  hypotenuse  and  an  acute  angle. 

34.  The  hypotenuse  and  the  altitude  upon  it. 

35.  The  median  and  the  altitude  upon  the  hypote- 
nuse.    [Same  as  No.  34.] 

36.  The  radius  of   the    circumscribed   circle  and 
a  leg. 

37.  The  radius  of  the  inscribed  circle  and  a  leg. 
Given:  Radius  =  r;    leg  =  CM.     Required:    (?). 

Analysis  :  Consider  that  A  BC  is  the  completed  figure ; 
CNOM  is  a  square,  whose  vertex  O  is  the  center  of 
the  circle,  and  side  OAris  the  given  radius.  AB  is  tan- 
gent from  A.  Construction  :  On  CA  take  CN '=  r  and 
construct  square,  CNOM.  Prolong  CM  indefinitely. 
BOBBINS'  PLANE  GEOM. — 9 


R    €..-• --\ 

r\\ 


C       H 

Describe  O,  etc. 


130 


PLANE   GEOMETRY 


38.  One  leg  and  the  altitude  upon  the  hypotenuse.' 

39.  An  acute  angle  and  the  sum  of  the  legs. 
Given:  AD  =  mm;   Z  K.    Required:    (?). 

Construction  :  At  A  construct  Z  A  —  Z  K;  at 
D  construct  Z  D  =  45°,  the  sides  of  these  A 
intersecting  at  B.  Draw  BC  ±  to  AD  ;  etc.  6 

40.  The  hypotenuse  and  the  sum  of  the  legs. 
[Use  A   as  center,  hypotenuse  as  radius,  etc.] 

41.  The  radius  of  the  circumscribed  circle  and  an  acute  angle. 

42.  The  radius  of  the  inscribed  circle  and  an  acute  angle. 
Construction:   Take    CS   on  indefinite  line  B 

ZA  =  r.  On  CS  construct  square  CSOM.  At 
0  construct  /.  MOX  =  Z  K.  Draw  radius  OT 


45". 


JL  to  OX.  Draw  tangent  at  T.     Proof  :  A  ABC 
is  a  rt.  A  and  it  is  the  rt.  A.    (Explain.) 


K 


III.   To  construct  an  equilateral  triangle,  having  given : 

43.  One  side. 

44.  The  altitude. 

45.  The  perimeter. 

46.  A  median. 

47.  The  radius  of  the  inscribed  circle. 

[Draw  circle  and  radius;  at  center  construct  Z.ROS 
=  120°  and  Z  ROT  =  120°;  etc.] 

48.  The  radius  of  the  circumscribed  circle. 


IV.   To  construct  a  triangle,  having  given  : 

49.  The  base,  an  angle  adjoining  it,  the  altitude  upon  it. 

50.  The  midpoints  of  the  three  sides. 
[Draw  R S,  R  T,  ST,  etc.] 

51.  One  side,  altitude  upon  it,  and  the  radius 

of  the  circumscribed  circle.  /     '*'••....  / 

Construction :    Draw  O  with  given  radius  and     A  ~"C 

any  center.     Take  chord  =  given  side  ;  etc. 

52.  One  side,  an  adjoining  angle,  and  the  radius  of  the  circumscribed 
circle. 

53.  Two  sides  and  the  altitude  from  the  same  vertex. 
Construction:    Erect  JL  —  altitude,  upon  an  indefinite  line.     Use  the 

end  of  this  altitude  as  center  and  the  given  sides  as  radii;  etc. 


BOOK  II  131 

54     One  side,  an   angle  adjoining  it,  and  the  D  /X 

sum  of  the  other  two  sides.  - 

Construction:  At  A  construct  Z  BA X  =  given  _  / 
£K.     On  AX  take  AD  =  s;    draw  DB ;    bisect 
DB  at  M,  etc. 

55.  Two  sides  and  the  median  to  the  third  side. 
Given :  a,  b,  m.    Construction :    Construct  A  ABR 

whose  three  sides,  AB  =  a,  BR  =  b,  AR  =  2  m. 
Draw^C  II  to  BR  and  RC  II  to  AB  meeting  at 
C.  Drawee.  Statement:  (?).  Proof:  (?)• 

56.  A  side,  the  altitude  upon  it,  and  the  angle 
opposite  it. 

Given :  Side  =AB,  alt.  =  h ;  opposite  Z  =  Z  C'. 

Construction:  Upon  AB  construct  segment 
ACB  which  will  contain  A  =  Z  C'  (by  276). 
At  A  erect  A R  JL  to  AB  and  =  A;  etc. 

57.  A  side,  the  median  to  it,  the  angle  oppo- 
site it. 

[Statement:  A  ABC  is  the  required  A.] 

58.  One  side  and  the  altitude  from  its  extremi 
ties  to  the  other  sides. 

Given:  Side  =  AB,  altitudes  x  and  y. 

Construction:  Bisect  AB;  describe  a  semicircle.  Using 
A  as  center  and  x  as  radius,  describe  arc  cutting  the 
semicircle  at  R;  etc. 

59.  Two  sides  and  the  altitude  upon  one  of  them. 
[Given  :  Sides  =  A  B  and  BC ;  alt.  on  EC  =  *.]  x 

60.  One  side,  and  angle  adjoining  it,  and  the  radius  of  the  inscribed 
circle. 

Construction:   Describe    O  with  given  radius,  any  center. 
Construct  central  Z  =  given  Z.     Draw  two  tangents  ||  to  these  radii. 


V.    To  construct  a  square,  having  given  : 

61.  One  side. 

62.  The  diagonal. 

63.  The  perimeter. 

64.  The  sum  of  a  diagonal  and  a  side. 


132  PLANE   GEOMETKY 

VI.  To  construct  a  rhombus,  having  given: 

65.  One  side  and  an  angle  adjoining  it. 

66.  One  side  and  the  altitude. 

67.  The  diagonals. 

68.  One  side  and  one  diagonal.     [Use  281.] 

69.  An  angle  and  the  diagonal  to  the  same  vertex. 

70.  An  angle  and  the  diagonal  between  two  other  vertices. 

71.  One  side  and  the  radius  of  the  inscribed  circle. 


VII.  To  construct  a  rectangle,  having  given: 

72.  Two  adjoining  sides. 

73.  A  diagonal  and  a  side. 

74.  One  side  and  the  angle  formed  by  the  diagonals. 

75.  A  diagonal  and  the  sum  of  two  adjoining  sides.     [See  No.  40.] 

76.  A  diagonal  and  the  perimeter. 

77.  The  perimeter  and  the  angle  formed  Y\  x 
by  the  diagonals.                                                                                            " 

Construction :  Bisect  the  perimeter  and 
take  A  B  =  half  it.    Bisect  Z  K.    At  A  con-  K 
struct  Z  BA X  =  half  Z  K.     Etc. 


VIII.    To  construct  a  parallelogram,  having  given: 

78.  One  side  and  the  diagonals.     [Use  137  and  278.] 

79.  The  diagonals  and  the  angle  between  them. 

80.  One  side,  an  angle,  and  the  diagonal  not  to  the  same  vertex. 

81.  One  side,  an  angle,  and  the  diagonal  to  the  same  vertex. 

82.  One  side,  an  angle,  and  the  altitude  upon  that  side. 

83.  Two  adjoining  sides  and  the  altitude. 


IX.  To  construct  an  isosceles  trapezoid,  having  given: 

84.  The  bases  and  an  angle  adjoining  the  larger  base. 

85.  The  bases  and  an  angle  adjoining  the  less  base. 

86.  The  bases  and  the  diagonal. 

87.  The  bases  and  the  altitude. 

88.  The  bases  and  one  of  the  equal  sides. 

89.  One  base,  an  angle  adjoining  it,  and  one  of  the  equal  sides. 

90.  One  base,  the  altitude,  and  one  of  the  equal  sides. 


BOOK   TT  133 

91.  One  base,  the  radius  of  the  circumscribed  circle,  and  one  of  the 
equal  sides.     [First,  describe  a  0.] 

92.  One  base,  an  angle  adjoining  it,  and  the  radius  of  the  circum- 
scribed circle. 

93.  The  bases  and  the  radius  of  the  circumscribed  circle. 

94.  One  base  and  the  radius  of  the  inscribed  circle. 
Construction :   Bisect  the  base  and  erect  a  _L  =  radius ;  etc. 


X.  To  construct  a  trapezoid,*  having  given: 

95.  The  bases  and  the  angles  adjoining  one  of  them. 
Construction:   Take  EC  =  longer  base,  and  on  it  take  ED  =  less  base. 

Construct  A  DEC  (by  280). 

96.  The  four  sides. 

97.  A  base,  the  altitude,  and  the  non-parallel  sides. 
Construction :   Construct  a  A  two  sides  of  which  =  the  given  non-11  sides 

of  the  trapezoid,  and  the  alt.  from  same  vertex  =  given  alt.     (See  No.  53.) 

98.  The  bases,  an  angle,  and  the  altitude. 

Construction:   Construct  O  on  ED,  having  given  altitude  and  Z. 

99.  A  base,  the  angles  adjoining  it,  and  the  altitude. 

100.  The  longer  base,  an  angle  adjoining  it,  and  the  non-parallel  sides. 

101.  The  shorter  base,  an  angle  not  adjoining  it,  and  the  non-parallel 
sides. 

XI.  To  construct  the  locus  of  a  point  which  will  be : 

102.  At  a  given  distance  from  a  given  point. 

103.  At  a  given  distance  from  a  given  line. 

104.  At  a  given  distance  from  a  given  circumference  : 
(i)  If  the  given  radius  is  <  the  given  distance ; 
(i»)  If  the  given  radius  is  >  the  given  distance. 

105.  Equally  distant  from  two  given  points. 

106.  Equally  distant  from  two  intersecting  lines. 


*  NOTE.     It  is  evident  that  every  trapezoid  may  be  divided  into  a 

parallelogram   and  a  triangle  by  drawing  one         A 

line  (as  BD)  II  to  one  of  the  non-11  sides.  Hence  / 
the  construction  of  a  trapezoid  is  often  merely  / 
constructing  a  triangle  and  a  parallelogram.  _' 


A 


134  PLANK   GEOMETRY 

XII.  To  find  (by  intersecting  loci)  *  the  point  P,  which  will  be : 

107.  At  two  given  distances  from  two  given  points. f 

108.  Equally  distant  from  three  given  points. 

109.  In  a  given  line  and  equally  distant  from  two  given  points. 

110.  In  a  given  line  and  equally  distant  from  two  given  intersecting 
lines. 

111.  In  a  given  circumference  and  equally  distant  from  two  given 
points,  f 

112.  In  a  given  circumference  and  equally  distant  from  two  inter- 
secting lines.f 

113.  Equally  distant  from  two  given  intersecting  lines  and  equally 
distant  from  two  given  points-! 

114.  At  a  given  distance  from  a  given  line  and  equally  distant  from 
two  given  points.f 

115.  At  a  given  distance  from  a  given  line  and  equally  distant  from 
two  other  intersecting  lines,  f 

116.  Equally  distant  from  two  given  points  and  at  a  given  distance 
from  one  of  them.f 

117.  Equally  distant  from  two  given  intersecting  lines  and  at  a  given 
distance  from  one  of  them.f 

118.  At  a  given  distance  from  a  point  and  equally  distant  from  two 
other  points.f 

119.  At  given  distances  from  two  given  intersecting  lines.f 

120.  At  given  distances  from  a  given  line  and  from  a  given  circum- 
ference.! 

121.  At  given  distances  from  a  given  line  and  from  a  given  point.f 

122.  Equally  distant  from  two  parallels  and  equally  distant  from 
two  intersecting  lines.f 

123.  At  a  given  distance  from  a  given  point  and  equally  distant  from 
two  given  parallels,  f 


*  It  is  well  to  draw  the  loci  concerned  as  dotted  lines.    (See  No.  124.) 
I  In  the  Discussion,  include  the  answers  to  questions  like  these  : 

(1)  Is  this  ever  impossible  ?  (i.e.  must  there  always  be  such  a  point  ?) 

(2)  Are  there  ever  two  such  points  ?    When  ? 

(3)  Are  there  ever  more  than  two  ?     When  ? 

(4)  Is  there  ever  only  one  ?    When  ?    Etc. 


BOOK  II  135 


124.  At  a  given  distance  from  a  given  point 
and  equally  distant  from  two  given  intersecting 
lines. 

Can  C  be  so  taken  that  there  will  be  no  point  ? 

Can  C  be  so  taken  that  there  will  be  only  one 
point?  Only  two?  Only  three?  More  than  four  ? 


XIII.    To  find  (by  intersecting  loci)  the  center  of  a  circle  which  will : 

125.  Pass  through  three  given  points.* 

126.  Pass  through  two  given  points  and  touch  a  given  line  at  a  given 
point.* 

127.  Pass  through  two  given  points  and  touch  two  given  intersecting 
lines.* 

128.  Have  a  given  radius,  touch  a  given  line,  and  pass  through  a 
given  point.* 

129.  Pass  through  a  given  point  and  touch  two  given  parallel  lines.* 

130.  Pass  through  two  given  points  and  touch  two  given  parallels.* 

131.  Have  a  given  radius  and  touch  two  given  intersecting  lines.* 

132.  Have  a  given  radius  and  pass  through  two  given  points.* 

133.  Touch  three  given    indefinite   lines,   no  two   of    them    being 
parallel.f 

134.  Touch  three  given  lines,  only  two  of  them  being  parallel. 


XIV.  To  construct  a  circle  which  will : 

135.  Pass  through  two  given  points  and  touch  a  given  line  at  a  given 
point. 

136.  Pass  through  two  given  points  and  touch  two  given  intersecting 
lines. 

137.  Pass  through  two  given  points  and  touch  two  given  parallels. 

138.  Have  a  given  radius,  touch  a  given  line,  and  pass  through  a 
given  point. 

139.  Have  its  center  in  one  line,  touch  another  line,  and  have  a  given 
radius. 


*  Discussion :    Is  this  ever  impossible  ?     Are  there  ever  two  circles  and 
hence  two  centers  ?     Are  there  ever  more  than  two  ?    Etc. 
t  Four  solutions.     One  is  in  274. 


136 


PLANE   GEOMETRY 


140.  Have  a  given  radius  and  touch  two  given  intersecting  lines. 

141.  Have  a  given  radius  and  pass  through  two  given  points. 

142.  Have  a  given  radius  and  touch  a  given  circumference  at  a  given 
point.     [Draw  tangent  to  the  given  O  at  the  given  point.] 

143.  Have  a  given  radius  and  touch  two  given  circumferences. 

144.  Touch  three  indefinite  intersecting  lines.* 

145.  Touch  two  given  intersecting  lines,  one  of  them  at  a  given  point. 

146.  Touch  a  given  line  and  a  given  cir- 
cumference at  a  given  point. 

Given  :    Line  AB  ;  O  C;  point  P. 

Construction :  Draw  radius  CP.  Draw 
tangent  at  P  meeting  AB  at  R.  Bisect 
Z.PRB,  meeting  CP  produced  at  O;  etc. 

147.  Be  inscribed  in  a  given  sector. 

Construction  :    Produce  the  radii  to  meet  the  tangent  at  the  midpoint 
of  the  arc.     In  this  A  inscribe  a  O. 

148.  Have  a  given  radius  and  touch  two  given  circles. 

149.  Have  a  given  radius,  touch  a  given  line,  and  a  given  circumference. 

•  150.  Touch  a  given  line  at  a  given  point 
and  touch  a  given  circumference. 

Given  :    Line  AB;   point  P;  O  C. 

Construction  :  At  P  erect  PX  ±  to  A  B,  and 
extend  it  below  AB,  so  PR  =  radius  of  O  C. 

Draw  CR  and  bisect  it  at  M. 

Erect  MY  ±  to  CR  at  M,  meeting  PX  at 
0;  etc. 


151.  What  is  the  locus  of  the  vertices  of  all  right  triangles  having 
the  same  hypotenuse? 

152.  Through  a  given  point  on  a  given  circumference  to  draw  two 
equal  chords  perpendicular  to  each  other. 

153.  To  draw  a  line  of  given  length  through  a  given  point  and  ter- 
minating in  two  given  parallels. 

Construction :  Use  any  point  of  one  of ,  the  Us  as  center  and  the 
given  length  as  radius  to  describe  an  arc  meeting  the  other  ||.  Join 
these  two  points.  Through  the  given  point  draw  a  line  II,  etc. 


*  Four  solutions.     One  is  in  274. 


BOOK  II 


137 


154.  To  draw  a  line,  terminating  in  the 
sides  of  an  angle,  which  shall  be  equal  to 
one  line  and  parallel  to  another. 

Statement  :   RS  is  =  a  and  ||  to  x. 


M 


155.  To  draw  a  line  through  a  given  point 

within  an  angle,  which  shall  be  terminated  by  the  sides  of  the  angle  and 
bisected  by  the  point. 

Construction:   Through  P  draw  PD 
II  to    AC.     Take    on    AB,   DE  =  AD. 

Draw  EPF-,  etc. 

^ 

156.  To  circumscribe  a  circle  about  a 

rectangle.  A  F         ^ 

157.  To  construct  three  circles  having  the  vertices  of  a  given  triangle 
as  centers  so  that  each  touches  the  other  two.  _^  c 

Construction  :   Inscribe  a  O  in  the  A  ;  etc. 

158.  To  construct  within    a   circle    three    equal 
circles  each  of  which  will  touch  the  given  circle  and 
the  other  two. 

Construction:   Draw  a  radius,  OA,  and  construct 
/.  A  OB  =  120°  and  Z  A  OC  =  120°.     In  these  sectors  inscribe,  etc. 

159.  Through  a  point  without  a  circle  to  draw 
a  secant  having  a  given  distance  from  the  center. 

160.  To  draw  a  diameter  to  a  circle  at  a  given 
distance  from  a  given  point. 

161.  Through  two  given  points  within  a  circle 
to  draw  two  equal  and  parallel  chords. 

Construction  :  Bisect  the  line  joining  the  given 
points  and  draw  a  diameter,  etc. 

162.  To  draw  a  parallel  to  side  RC  of  tri- 
angle ABC,  meeting  AB  in  X  and  ^ICin   F, 
such  that  XY=YC. 


163.  Find  the  locus  of  the  points  of  contact 
of  the  tangents  drawn  to  a  series  of  concentric 
circles  from  an  external  point. 

164.  Given:  Line  AB  and  points  C  and  D 
on  the  same  side  of  it;  find  point  X  in   AB 

such  that  Z  A  XC  =  Z  BXD.  \  /" 

Construction:    Draw  CE±io  AB  and  pro-  F 

duce  to  F  so  that  EF  =  CE.     Draw  FD  meeting  AB  in  X. 


Draw  CX. 


138 


PLANE   GEOMETRY 


165.  To  drawr  from  one  given  point  to  another 
the  shortest  path  which  shall  have  one  ^)oint  in 
common  with  a  given  line. 

Statement :  CX  +  XD  is  <  CR  +  RD. 

166.  To  draw  a  line  parallel  to  side  BC  of  tri- 
angle ABC  meeting  AB  at  X  and  AC  at  F,  so 
tha,tXY=BX  +  YC. 

Construction :  Draw  bisectors  of  A  B  and  C,  meeting  at  O,  etc. 

167.  To  draw  in  a  circle,  through  a  given  point  of  an  arc,  a  chord 
which  will  be  bisected  by  the  chord  of  the  arc. 

Construction  :  Draw  radius  OP  meeting  chord  at  C. 
Prolong  PO  to  X  so  CX  =  CP.     Draw  XM  II  to  AB 
meeting  O  at  M.     Draw  PM  cutting  AB  at  Z>;  etc.      A> 
Is  there  any  other  chord  from  P  bisected  by  AB7 

168.  To  inscribe  in  a  given  circle  a  triangle  whose  angles  are  given. 
Construction :    At    the    center    construct   three  A,  doubles   of   the 

given  A. 

169.  To  circumscribe  about  a  given  circle  a  triangle  whose  angles  are 
given. 

Construction :   Inscribe  A  (like  No.  168)  first,  and  draw  tangents  II  to 
the  sides. 

170.  Three  lines  meet  in  a  point ;  it  is  required 
to  draw  a  line  terminating  in  the  outer  two  and 
bisected  by  the  inner  one. 

Construction :  Through  any  point  P,  of  OB, 
draw  Us  to  the  outer  lines.  Draw  diagonal  RS;  etc. 

171.  To  draw  through  a  given  point,  P,  a  line 
which  will  be  terminated  by  a  given  circumference 
and  a  given  line  and  be  bisected  by  P. 

Construction:  Draw  any  line  DX  meeting  AB 
at  D.  Draw  PE  It  to  AB  meeting  DX  at  E.  Take 
EF=ED;  etc. 

172.  Through  a  given  point  without  a  circle  to 
draw  a  secant  to  the  circle  which  shall  be  bisected 
by  the  circumference. 

Construction  :  Draw  arc  at  T,  using  P  as  center 
and  diam.  of  O  0  as  radius.  Using  T  as  center  and 
same  radius  as  before,  describe  circumference  touch- 
ing O  0  at  C  and  passing  through  P.  Draw  PC 
meeting  O  O  at  M. 


ROOK    II 


139 


173.  To  inscribe  a  square  in  a  given  rhombus. 
[Bisect  the  four  A  formed  by  the  diagonals.] 

174.  To  bisect  the  angle  formed  by  two 
lines  without  producing  them  to  their  point 
of  intersection. 

Construction :  At  P,  any  point  in  RS, 
draw  PA  II  to  XY-,  bisect  Z  APS  by  PB. 
At  any  point  in  PB  erect  ML  J_  to  PB, 
meeting  the  given  lines  in  M  and  L.  Bisect 
ML  at  D  and  erect  DC  JL  to  ML,  etc. 

175.  To  construct  a  common  external  tangent  to  two  circles. 
Construction :  Using  0  as  a  center  and  a  B 

radius  =  difference  of  the  given  radii,  con- 
struct (dotted)  circle.  Draw  QA  tangent  to 
this  O  from  Q ;  draw  radius  OA  and  produce 
it  to  meet  given  O  at  B.  Draw  radius  QC  \\ 
to  OB.  Join  BC. 

Statement :  BC  is  tangent  to  both  (D. 

Proof :   AB  =  CQ  (Const.).     AB  is  ||  to  CQ  (?). 

.'.ABCQisa.  O  (?). 

But  Z  OA  Q  is  a  rt.  Z  (?) ;  etc. 

176.  To  construct  a  common  internal 
tangent  to  two  circles. 

Construction :  Using  0  as  a  center  and 
a  radius  =  the  sum  of  the  given  radii, 
construct  (dotted)  circle.  Draw  QA  tan- 
gent to  this  O  from  Q ;  draw  radius  OA 
meeting  given  O  at  B,  etc.,  as  above. 


BOOK  III 

PROPORTION.    SIMILAR  FIGURES 

283.  A  ratio  is  the  quotient  of  one  quantity  divided  by 
another, — both  being  of  the  same  kind. 

284.  A  proportion  is  the  statement  that  two  ratios  are  equal. 

285.  The  extremes  of  a  proportion  are  the  first  and  last 
terms. 

The  means  of  a  proportion  are  the  second  and  third  terms. 

286.  The  antecedents  are  the  first  and  third  terms. 
The  consequents  are  the  second  and  fourth  terms. 

287.  A  mean  proportional  is  the  second  or  third  term  of  a 
proportion  in  which  the  means  are  identical. 

A  third  proportional  is  the  last  term  of  a  proportion  in 
which  the  means  are  identical. 

A  fourth  proportional  is  the  last  term  of  a  proportion  in 
which  the  means  are  not  identical. 

288.  A  series  of  equal  ratios  is  the  equality  of  more  than 
two  ratios. 

A  continued  proportion  is  a  series  of  equal  ratios  in  which 
the  consequent  of  any  ratio  is  the  antecedent  of  the  next 
following  ratio. 

289.  EXPLANATORY.     A  ratio  is  written  as  a  fraction  or  as  an  indi- 
cated  division ;   -,  or  a  -H  6,  or   a :  b.     A  proportion   is   usually  written 

7-  =  -»  or  a  :  b  —  x  :  y,  and  is  read  :  "  a  is  to  b  as  x  is  to  y"     In  this  pro- 
y 

portion  the  extremes  are  a  and  y ;  the  means  are  b  and  x ;  the  antece- 
dents are  a  and  x ;  the  consequents  are  b  and  y ;  and  y  is  a  fourth 
proportional  to  a,  6,  x.  In  the  proportion  a  :  m  =  m  :  z,  the  mean  pro- 
portional is  m,  and  the  third  proportional  is  z. 

140 


BOOK   III  141 

THEOREMS    AND    DEMONSTRATIONS 

290.  THEOREM.   In  a  proportion  the  product  of  the  extremes  is 
equal  to  the  product  of  the  means. 

Given  :  ^  =  —  or  a  :  b  =  x  :  y.     To  Prove  :  ay  =  bx. 
b       y 

Proof:  —  =  —  (Hyp.).     Multiply  by  the  common  denomi- 
b      y 

nator,  by  and  obtain,  ay  =  bx  (Ax.  3).  Q.E.D. 

291.  THEOREM.    If  the  product  of  two  quantities  is  equal  to  the 
product  of  two  others,  one  pair  may  be  made  the  extremes  of  a  propor- 
tion and  the  other  pair  the  means. 

Given  :  ay  —  bx. 

To  Prove  :  These  eight  proportions : 

1.  aib  =  x:y,  5.  xiy  =  a:b, 

2.  a:x=b:y,  6.  x:a  =  y:b, 

3.  b  :  a  =  y  :  x,  7.  y  :  x  =  b  :  a, 

4.  b  :  y  =  a  :  x,  8.  y  :  b  —  x  :  a. 

Proof:  1.  ay  =  bx    (Hyp.).     Divide  each   member  by  by, 

and  obtain  -^  =  -^  (Ax.  3).     .-.  -^  =  -,  or  a :  b  =  x :  y.  Q.E.D. 
by      by  by 

2.  Divide  by  xy\  etc.    3.  bx=ay(Hyp.').  Divide  by  ax;  etc. 

NUMERICAL  ILLUSTRATION.  Suppose  in  this  paragraph  a  =  4,  b  =  14, 
x  =  6,  y  =  21 ;  the  truth  of  the  above  proportions  can  be  clearly  seen  by 
writing  these  equivalents.  4  x  21  =  14  x  6  (True). 

1.    4  : 14  =  6  :  21  (True) ;  2.  4  :  6  =  14  : 21  (True)  ;  etc. 

They  will  all  be  recognized  as  true  proportions. 

292.  THEOREM.    In  any  proportion  the  terms  are  also  in  proportion 
by  alternation  (that  is,  the  first  term  is  to  the  third  as  the  second 
is  to  the  fourth). 

Given  :  a  :  b  =  x  :  y.     To  Prove  :  a:x—b\y. 

Proof:  a:b  =  x:y   (Hyp.).      .-.    ay  =  bx   (290). 

Hence,  a:x=b:  y  (291).  Q.E.D. 


142  PLANE   GEOMETRY 

293.  THEOREM.   In  any  proportion  the  terms  are  also  in  proportion 
by  inversion  (that  is,  the  second  terra  is  to  the  first  as  the  fourth 
term  is  to  the  third). 

[The  proof  is  similar  to  the  proof  of  292.] 

294.  THEOREM.    In  any  proportion  the  terms  are  also  in  proportion 
by  composition  (that  is,  the  sum  of  the  first  two  terms  is  to  the 
first,  or  second,  as  the  sum  of  the  last  two  terms  is  to  the  third, 
or  fourth). 

(  a  +  b:  a  =  x-\-y:z,  or 
Given:  a:  b  =  x:  y.       To  Prove  i\        ,    , 


Proof:   a:  b  =  x:y  (Hyp.).    .'.  ay  =  bx  (?)  (290). 

Add  ax  to  each,  and  obtain,   ax  +  ay  —  ax  +  bx   (Ax.  2). 

That  is,  a  (x  +  y)  =  x  (  a  +  £). 

Hence,  a  +  b  :  a  =  x  +  y  :  x  (?)  (291). 

Similarly,  by  adding  by,  a  +  b  :  b  =  x  +  y  :  y.  Q.E.D. 

295.  THEOREM.  In  any  proportion  the  terms  are  also  in  proportion 
by  division  (that  is,  the  difference  between  the  first  two  terms  is  to 
the  first,  or  second,  as  the  difference  between  the  last  two  terms 
is  to  the  third,  or  fourth). 

(a  —  b  :  a  =  x—  y  '.  x,  or 
Given  :  a  :  b  =  x  :  y.     To  Prove  :  \         ,     , 

\a-b:b  =  x-yiy. 

Proof:   a:b  =  xiy    (Hyp.).     .-.  ay  =  5*  (?)   (290). 

Subtracting  each  side  from  ax,  ax—  ay  —  ax—  bx  (Ax.  2). 

That  is,  a(x  —  y)  =  x(a  —  6). 

Hence,  a-b  :  a  =  x-  y  :  x  (?)  (291). 

Likewise,  a  —  b  :b  =  x—  y  :  y.  Q.E.D. 


NOTE  I.    The  proportions  of  294  and  295  may  be  written  in  many 
different  forms  (292,  293) .     Thus,  (1)  a  ±b:a  =  x±y:  x; 
(2)  a  ±  b  :  b  =  x  ±  y  :  y ;   (3)  a  ±  b  :  x  ±  y  =  a  :  x,  etc. 

NOTE  II.  In  any  proportion  the  sum  of  the  antecedents  is  to  the  sum 
of  the  consequents  as  either  antecedent  is  to  its  consequent.  (Explain.) 
Also,  in  any  proportion  the  difference  of  the  antecedents  is  to  the  differ- 
ence of  the  consequents  as  either  antecedent  is  to  its  consequent. 
(Explain.)  Thus:  o  +  x:b+  y  =  a:b  =  x  :  y. 

Also,  a-x :b  —  y  —  a:b  =  x:y. 


BOOK    TTT  148 

296.  THEOREM.    In  any  proportion  the  terms  are  also  in  proportion 
by  composition  and  division  (that  is,  the  sum  of  the  first  two  terms 
is  to  their  difference  as  the  sum  of  the  last  two  terms  is  to  their 
difference). 

Given :   a  :  b  =  x :  y.     To  Prove  :  a+  ,  =  x  "*"^- 

a  —  b     x—  y 

ft  -I     rk         /y*  —I.  #/ 

Proof:  a_±?.  =  x_±y_  (?)  (294). 
a  x 

aL-^==x--jt  ^  ^95^ 

Divide  the  first  by  the  second,  a      _  =x     ^  (?).      Q.E.D. 

a  —  b     x  —  y 

297.  THEOREM.    In  any  proportion,  like  powers  of  the  terms  are 
also  in  proportion,  and  like  roots  of  the  terms  are  in  proportion. 

Given  :    a  :  b  =  x  :  y. 

To  Prove :    an  :  bn  =  xn  :  yn ;   and  Sfa  :  3/b  =  %/x :  ^y. 

Proof :   [Write  the  given  proportion  in  fractional  form,  etc.] 

298.  THEOREM.   In  two  or  more  proportions  the  products  of  the  cor- 
responding terms  are  also  in  proportion. 

Given  :    aib  =  x:y,  and  c :  d  =  I :  m,  and  e  :/=  r :  8. 

To  Prove  :    ace  :  bdf=  xlr :  yms. 

Proof:    [Write  in  fractional  form  and  multiply.] 

299.  THEOREM.    A  mean  proportional  is  equal  to  the  square  root  of 
the  product  of  the  extremes. 

Given :   a :  x  —  x :  b.     To  Prove :  x  =  Va6. 
Proof :    [Use  290 ;  etc.] 

300.  THEOREM.    If  three  terms  of  one  proportion  are  equal  to  the 
corresponding  three  terms  of  another  proportion,  each  to  each,  the  re- 
maining terms  are  also  equal. 

f  a  :  b  =  c  :  m,  and  ) 
Given :  v      ,  K     To  Prove :  m  =  r. 

(  a :  b  =  c :  r.  } 

Proof :    am  =  Jc.and  ar  =  be  (?)  (290). 

.-.  am  =  ar  (Ax.  1).     Hence,   m  =  r  (Ax.  3).  Q.E.D. 


144  PLANE   (IKOMETRY 

301.  THEOREM.  In  a  series  of  equal  ratios,  the  sum  of  all  the  ante- 
cedents is  to  the  sum  of  all  the  consequents  as  any  antecedent  is  to  its 
consequent. 

Given:    f  =  £  =  Hi- 

b      d     f     h 

To  Prove:    £±±±£±4=  f  =  l  -etc. 

b  +  d  +f+  h      b      d 

Proof  :    Set  each  given  ratio  =  m  ;   thus, 

a  c  e  a 

l  =  m\  -=mi  -  =  w;f  =  w. 

b  d  f  h 

.'.  a=bm,  c  =  dm,  e=fm,  g  =  hm  (Ax.  3). 


Hence   a  +  c  +e+g     bm  +  dm+fm  +  hm    (Substitution) 
l+d+f+h  b+d+f+h 


=  m  (Canceling). 


'  b  +  d+f+h      b      d     f     h  Q-B.D. 

EXERCISES 

1.  If  3:4  =  6:z,  find  x.  2.   If  8  :  12  =  12  :  ar,  find  x. 

3.  Find  a  fourth  proportional  to  6,  7,  and  15. 

4.  Find  a  third  proportional  to  4  and  10. 

5.  If  11:  15  =  x:  25,  find  x.  6.    If  4  :  x  =  x  :  25,  find  x. 

7.  Find  a  mean  proportional  between  8  and  18. 

8.  If  7:  x  =  35:  48,  find  x. 

9.  Given,  that  5:8=15:  24,  write  seven  other  true  proportions  con- 
taining these  same  four  numbers. 

10.  If  5x6  =  2x1  5,  write  eight  proportions  with  these  numbers. 

11.  If  7  :  12  =  21  :  36,  write  the  proportion  resulting  by  alternation  ; 
inversion  ;  composition  ;  division  ;  composition  and  division. 

12.  If  6  :  25  =  18  :  75,  write  the  proportions  required  in  No.  11. 

13.  li  x  +  y:x  —  #=17:7,  write  the  proportions  that  result  by  virtue 
of  composition  ;  division  ;  composition  and  division. 

14.  Apply  301  to  the  ratios,  f  =  f  =  f  =  T\  =  ft. 


BOOK  TIT  145 

NOTE.  We  have  seen  that  it  is  possible  to  add  two  lines  and  subtract 
one  line  from  another.  Now  it  is  essential  that  we  clearly  understand 
the  significance  implied  by  indicating  the  multiplication  or  the  division 
of  one  line  by  another. 

What  is  actually  done  is  to  multiply  or  divide  the  numerical  measure 
of  one  line  by  the  numerical  measure  of  another.  Thus  if  one  line  is  8 
inches  long  and  another  is  18  inches  long,  we  say  that  the  ratio  of  the 
first  line  to  the  second  is  T\  or  f,  meaning  that  the  less  line  is  four  ninths 
of  the  larger. 

Also,  in  referring  to  the  product  of  two  lines  we  merely  understand 
that  the  product  of  their  numerical  measures  is  intended. 

If  a  line  is  multiplied  by  itself,  we  obtain  the  square  of  the  numerical 
measure  of  the  line.  The  square  of  the  line  AB  is  written  AB2  or 
(AB)'2  and  the  quantity  that  is  squared  is  the  numerical  value  of  the 
length  of  A  B. 

In  the  preceding  paragraphs  of  Book  TIT,  we  have  been  considering 
numerical  magnitudes.  It  should  be  distinctly  understood  that  in  the 
following  geometrical  propositions  and  demonstrations,  the  foregoing- 
interpretation  is  implied  in  multiplication  and  division  involving  lines. 

302.  THEOREM.  A  line  parallel  to  one  side  of  a  triangle  divides  the 
other  sides  into  proportional  segments. 

Given  :    A  ABC  and  line  LM  A 

II  to  BC. 

To  Prove: 

AL  :  LB  =  AM  :  MC. 

Proof:  I.  If    the    parts   AL 
and  LB  are  commensurable. 

There  exists  a  common  unit  _  __  _  _ 

of  measure  of  AL  and  LB  (238).  E 
Suppose  this  is  contained  9  times  in  AL  and  5  times  in  LB. 

Then,  ||=  |  (Ax.  3). 

LiB        O 

Draw  lines  through  the  several  points  of  division  II  to  BC. 
These  will  divide  AM  into  9  parts  and  MC  into  5  parts. 
All  of  these  14  parts  are  equal  (?)  (147). 


Hence,         =      (Ax.  3).        .-.        =        (Ax.  1). 

MC       5  LB       MC  Q.E.U. 

BOBBINS'  PLANE  GEOM.  —  10 


146  PLANE    GEOMETRY 

II.  If  the  parts  AL  and  LB  are 
incommensurable  . 

There  does  not  exist  a  common 
unit  (238).  Divide  AL  into  sev- 
eral equal  parts  (by  271).  Ap- 
ply one  of  these  as  a  unit  of 
measure  to  LB.  There  will  be  a  _ 

B  -  C 

.remainder,  #B,  left  over  (238). 
Draw  RS   ||   to  BC. 

Now  —  =  —  (The  commensurable  case). 
LR     MS 

Indefinitely  increase   the   number   of    equal  parts  of  AL. 
That  is,  indefinitely  decrease  each  part,  the  unit  or  divisor. 
Hence,  the   remainder,  RB,  will  be  indefinitely   decreased. 
(Because  the  remainder  is  <  the  divisor.) 
That  is,  RB  will  approach  zero  as  a  limit, 
and  SC  will  approach  zero  as  a  limit. 

/.  LR  will  approach  LB  as  a  limit    (240), 
and  MS  will  approach  MC  as  a  limit   (240). 

.-.  —  will  approach  —  as  a  limit  (243), 

LR  LB 


and  -  will  approach  —^—  as  a  limit    (243). 
MS  MC 


Consequently,         =         (?)  (242).  Q.E.D. 

LB       MC 

303.  THEOREM.  If  a  line  parallel  to  one  side  of  a  triangle  intersects 
the  other  sides,  these  sides  and  their  corresponding  segments  are 
proportional. 

Given  :  A  ABC  ;  LM  \\  to  BC. 

To  Prove  : 
I.   AB  :  AC  =  AL  :  AM. 

II.    AB:  AC=  LB  :  MC. 
Proof  : 

AL  :  LB  =  AM:  MC  (?)  (302). 


BOOK  III  147 

/.     I.    AL  +  LB  :  AL  =  AM  +  MC  :  AM  (?)  (294), 
and    II.    AL  +  LB  :  LB  =  AM  +  M C  :  M C  (?). 

But  AL  +  LB  =  AB  and  AM  +  MC  =  AC  (Ax.  4). 

Therefore,  I.  ^L#  :  AL  =  AC:  AM  (Ax.  6). 

.Hence,  AB  :  AC  =  AL  :  AM  (?)  (292). 

II.   AB:LB  =  AC:MC  (Ax.  6). 

Hence,  AB  :  AC  =  LB  :  MC  (?)  (292).  Q.E.D. 

NOTE.   Each  of  these  proportions  may  be  written  eight  ways   (291). 

And  they  may  be  combined,  thus,  —  =  —  =  —  (Ax.  1). 

A  C      A  M.      Jvl  C 

304.  Two  lines  are  divided  proportionally,  if  the  ratio  of 
the  lines  is  equal  to  the  ratios  of  corresponding  segments. 

305.  THEOREM.   If  a  line  parallel  to  one  side  of  a  triangle  intersects 
the  other  sides,  it  divides  these  sides  proportionally. 

(Because  the  ratio  of  the  sides  =  the  ratio  of  correspond- 
ing segments  (303).     This  theorem  is  the  same  as  303.) 

306.  THEOREM.    Three  or  more  parallels  intercept  proportional  seg- 
ments on  two  transversals. 

Given:   (?). 

To   Prove  :    AC :  BD  =  CE :  DF 
=  EG :  FH. 

Proof:  Draw  from  A,  AT  II  to 
BH  intersecting  the  11$,  etc. 


In  A  AES,  /  \  \ 

^=^£  =  ^(?)  (305).  /  \ 

AS       AR       ES 

In  A  AGT,  4E  =  E^  (?)  (305). 
AS       ST 

.    AC  _CE_EG    ,A       -,, 
.  . »  XXA.    xi. 

AR       ES       ST 

But,  AR  =  BD,  RS=DF,    <ST=  FH  (?)  (130). 
Hence,  AC:  ED—  CE:  DF=  EG  :  FIT  (Ax.  6).  Q.E.D. 


148 


PLANE    GEOMETRY 


307.  THEOREM.    If  a  line  divides  two  sides  of  a  triangle  proportion- 
ally, it  is  parallel  to  the  third  side. 

Given:    A  ABC\    line  DE\    the 
proportion  AB  :  AC  —  AT) :  AE. 

To  Prove :  DE  is  II  to  BC. 

Proof :   Through  D  draw  DX  II 
to  BC  meeting  AC  at  X. 

.'.  AB  :  AC=AD  :  AX  (?)  (305). 
But  AB  :  AC  =  AD  :  AE  (Hyp.). 
.'.AX=AE(1)  (300). 

.-.  DX  and  DE  coincide  (?)  (39).     That  is,  DE  is  II  to  BC. 

CJ.E.D. 

308.  THEOREM.    The  bisector  of  an  angle  of  a  triangle  divides  the 
opposite  side  into  segments  which  are  proportional  to  the  other  two 
sides. 

Given :  A  ABC ;    BS  the    bi-   PN..% 
sector  of  Z  ABC.  \      ''"*••.._ 

To  Prove  :   AS  :  SC  =  AB  :  BC. 

Proof  :  Through  A  draw  AP  II 
to  BS,  meeting  BC,  produced, 
at  P. 

Now,  Z  m  =  Z  x  (?)  (98)  and  Z  n  =  Z  z  (?). 

But  Z  m  =  ^n  (Hyp.). 

.-.  Z  z  =  Z  z  (Ax.  1).     Hence  ^B  =  BP  (?)  (120). 

In  A  CAP,  BS  is  II  to  AP  (Const.). 

.'.AS:8C=BP:BC  (?)  (302). 

.'.  AS:  SC=AB:  BC  (Ax.  6). 


I.E.D. 


Ex. 
Ex. 


If  AS  = 


=  4,  £C  =  9,  find  SC. 
=9,BC  =  21,  find  A  S  and  SC. 


309.  The  segments  of  a  line,  made  by  one  of  its  points,  are 
the  lines  between  this  point  and  the  extremities  of  the  line. 

Thus,  in  308,  the  point  S  divides  AC  internally,  into  seg- 
ments SA  "and  SC. 


BOOK   TIT 


149 


But,  if  the  point  s'  be  in  the  prolongation  of  the  given  line, 
the  segments  are  still  s'A  and  s'c,  according  to  the  defini- 
tion, and  the  point  8f  di- 
vides AC,  externally,  into  s' £ 

segments  S1  A  and  SfC. 

310.  THEOREM.  The  bisector  of  an  exterior  angle  of  a  triangle  di- 
vides the  opposite  side  (externally)  into  segments  which  are  propor- 
tional to  the  other  two  sides. 

Given :  A  ABC;  BSf,  the  bisector  of  exterior  Z  ABD,  meet- 
ing AC  (externally)  at  sr.  To  Prove:  AS1  :  SfC  =  AB  :  BC. 

Proof :  Through  A  draw  AP  II  to  BSr  meeting  BC  at  P. 

Now,  Z  m  =  Z  x  (?). 

Z  n  =  /_  z  (?). 

But  Z  m  =  Z  n  (?). 

.-.Z  z  =  Zz  (?). 

Hence,  AB  =  BP  (?).  In 
A  CBS',  AP  is  II  to  BS'  (?). 
.'.  AS':  SfC  =  BP:  J3C(?)  (305). 
.'.  AS1  :  SfC=AB  :  BC  (?). 


Q.E.D. 


Ex.     If  AS'  =  10,  AB  =  7,BC  =  16,  find  S'C  and  AC. 
Ex.     If  ^  C  =  14,  4B  =  12,  £C  =  19,  find  A  S'  and  S'C. 


3UL   A  line  is  divided  harmonically  if  it  is  divided  in- 
ternally and  externally  in  the  same  ratio. 

In  308,  the  line  A  C  is  divided  internally  by  S,  in  the  ratio  AB :  BC. 
In  310,  the  line  A  C  is  divided  externally  by  S'  in  the  ratio  AB:  BC. 

THEOREM.    The  bisectors  of  the  interior  and  exterior  angles  of  a  tri- 
angle (at  a  vertex)  divide  the  opposite  side  harmonically. 

Given:  A  ABC;  BS  bisecting  Z.ABC;  and  BS'  bisecting  Z  ABD. 
To  Prove:   AS:  SC  =  AS'i  S'C.  D' 

*«*  41=  41  (*><*»>• 


C     BC^ 

Jc~  We 


Q.E.D. 


150 


PLANE   GEOMETRY 


312.  Similar   polygons   are   polygons   that   are    mutually 
equiangular   and  whose  homologous  sides 

are  proportional.  That  is,  every  pair  of 
homologous  angles  are  equal ;  and  the  ratio 
of  one  pair  of  homologous  sides  is  equal  to 
the  ratio  of  every  other  pair  of  homologous 
sides,  a  :  a'  —  b  :  V  =  c :  c'  =  d :  d'  =  etc. 

Triangles  are  similar  if  they  are  mutually  equiangular  and 
their  homologous  sides  are  proportional. 

313.  THEOREM.    Two  triangles  are  similar  if  they  are  mutually 
equiangular. 

Given :  A  ABC,  DEF-,  Z  ^i  =  Z  D,  Z  #  =  Z  #,  Z  c  =  Z>. 

To  Prove :  The  A  are  similar  (that  is,  that  their  sides  are 
proportional). 


Proof :  Place  A  ABC  upon  A  DEF  so  that  Z  A  coincides 
with  its  equal,  Z  D,  and  A  ABC  takes  the  position  of  A  DBS. 

Then  Z  DRS  =  ^E  (Hyp.).     .-.  BS  is  II  to  EF  (?)  (102). 

Hence,  DE  :  DB  =  DF  :  DS  (?)  (305). 

That  is,  DE  :  AB  =  DF :  AC  (Ax.  6). 

Likewise,  by  placing  Z  B  upon  Z  E,  we  may  prove  that, 
DE  :  AB  =  EF:  EC. 

.'.  DE  :  AB  =  DF:  AC  =  EF  :  BC  (Ax.  1). 

Therefore,  the  A  are  similar  (?)  (312).  Q.E.D. 


BOOK   III 


151 


314.  THEOREM.    Two  triangles  are  similar  if  two  angles  of  one  are 
equal  to  two  angles  of  the  other.     [See  117  and  313.] 

315.  THEOREM.    Two  right  triangles  are  similar  if  an  acute  angle 
of  one  is  equal  to  an  acute  angle  of  the  other.    [See  314.] 

316.  THEOREM.   If  a  line  parallel  to  one  side  of  a  triangle  intersects 
the  other  sides,  the  triangle  formed  is  similar  to  the  original  triangle. 

Given :  MN  II  to  BCin  A  ABC. 

To  Prove :  A  AMN  similar 
to  A  ABC. 

Proof :  Z  A  is  common  to 
both;  Z^L¥#  =  ZJ3;  Z  ANN 
=  Z  C  (?)  (98). 

/.  A  are  similar  (?)  (313). 

Q.E.D. 


317.  THEOREM.  If  two  triangles  have  an  angle  of  one  equal  to  an 
angle  of  the  other  and  the  sides  including  these  angles  proportional,  the 
triangles  are  similar. 


B  C        E  F 

Given  :   A  ABC  and  DEF\  Z  A  =  Z  D;   DE:AB  =  DF:AC. 
To  Prove:  The  A  similar. 

Proof :   Superpose  A  ABC  upon  A  DEF  so  that  Z  A  coincides 
with  its  equal,  Z  D,  and  A  ABC  takes  the  position  of  A  DBS. 
Then  DE:  DR  =  DF:  DS  (Hyp.).    /.  BS  is  II  to  J£F(?)  (307). 
/.  A  DBS  is  similar  to  A  DEF  (?)  (316).  Q.E.I). 


152  PLANE   GEOMETRY 

318.   THEOREM.   If  two  triangles  have  their  homologous  sides  pro- 
portional, they  are  similar. 

b 


B  C          E  F 

Given  :   A  ABC  and  DEF,  and  DE :  AB  =  DF:  AC  =  EF  :  BC. 
To  Prove :  A  ABC  similar  to  A  DEF. 

Proof:   On  DE  take  DK—AB\  and  on  DF  take  DL  =  AC. 
Draw  KL. 

Now,    DE  :  AB  =  DF:  AC  (Hyp.)- 

.'.  DE  :  DK  =  DF  :  DL  (Ax.  6).     .'.  KL  is  I!  to  EF  (?)  (307). 

Therefore,  A  DKL  is  similar  to  A  DEF  (?)  (316). 
[Now  A  ABC  is  to  be  proven  equal  to  A  YX7TL.] 

DE:DK=EF:KL.    (Definition  of  similar  triangles,  312.) 

That  is,  DE :  AB  =  EF  :  KL  (Ax.  6). 

But,  DE:  AB  =  EF:  BC  (Hyp.).      .'.  BC  =  KL  (300). 

Hence,  A  ABC  =  A~DKL  (?)  (58). 

But  A  DKL  has  been  proven  similar  to  A  DEF. 

Therefore,  A  ABC  is  similar  to  A  DEF  (Ax.  6).          Q.E.D. 


Ex.  1.    Are  all  equilateral  triangles  similar?     Why? 

Ex.2.    Are  all  squares  similar?     Why? 

Ex.  3.    Are  all  rectangles  similar?     Why? 

Ex.  4.  The  sides  of  a  triangle  are  7,  8,  and  12,  and  the  longest  side 
in  a  similar  triangle  is  30.  Find  the  other  sides. 

Ex.  5.  In  the  figure  of  311.  if  AB  =  10,  AC  =  14,  BC  =  18,  find  the 
four  segments  of  AC  made  by  S  and  S'. 

Ex.  0.    Prove  the  theorems  of  142  and  143  by  proportion. 


BOOK   III  153 

319.    THEOREM.     If   two  triangles  have   their   homologous    sides 
parallel,  they  are  similar. 

A  D 


E 

Given:   A  ABC  and  DEF;    AB  II  to  DE\    AC  II  to  DF;   and 
BC  II  to  EF. 

To  Prove  :  A  ABC  similar  to  A  DEF. 

Proof:  Produce  BC  of  A  ABC  until  it  intersects  two  sides 
of  A  DEF  at  R  and  8. 

Now  Z  B  =  Z  Z>#S,  and  Z  ZXRS  =  Z  E(?)  (98). 


Likewise,  Z.ACB  =  Z  D.S#,  and  /.DSR  =  /_  F  (?). 


Therefore,  A  ABC  is  similar  to  A  DEF  (?)  (314).       Q.E.D. 

320.   THEOREM.    If  two  triangles  are  similar  to  the  same  triangle, 
they  are  similar  to  each  other. 

Proof:   The  three  angles  of  each  of  the  first  two  triangles 
are  respectively  equal  to  the  three  angles  of  the  third  (312). 
Hence,  the  first  two  A  are  mutually  equiangular  (Ax.  1). 
Therefore  they  are  similar  (?)  (313).  Q.E.D. 


Ex.  1.  Let  the  pupil  prove  the  theorem  of  #19  if  one  triangle  entirely 
surrounds  the  other. 

Ex.  2.    If  one  side  of  one  triangle  intersects  two  sides  of  the  other. 

Ex.  3.  If  they  are  so  placed  that  no  side  of  either,  when  prolonged, 
intersects  any  side  of  the  other  without  being  prolonged. 

[Prolong  any  side  of  one  and  the  sides  not  ||  to  it  in  the  other.] 


154 


PLANE   GEOMETRY 


321.   THEOREM.    If  two  triangles  have  their  homologous  sides  per- 
pendicular, they  are  similar. 


c 

Given:  A  ABC  and  DEF-,  AB  _L  to  DE-,  AC  A.  to  DF-, 
BC  J_  to  EF. 

To  Prove :  A  ABC  similar  to  A  DEF. 

Proof :  Through  P,  any  point  in  EF,  construct  PR  II  to  AC, 
meeting  DF  at  M.  At  R,  any  point  in  PM,  draw  RS  II  to  AB, 
meeting  ED  at  N.  Draw  PS  II  to  BC,  meeting  NR  at  S,  forming 
the  APRS.  PJfisJ_toDFand  #^is_L  toZ)j£(?).  In  quadrilat- 
eral D3f£#,  Z  D  +  Z  Jf  +  Z  Jffijy  +  Z  2V  =  4  rt.  A  (?)  (165). 

But,  Z  3f +^  JT=2  rt.  ^  (Const.). 

/.  Z  D  +  Z  J/flJvr  =  2  rt.  Zs  (Ax.  2). 

But,  Z  a  +  Z  MRN=2  rt.  Zs  (?)  (46). 

.-.  Z  D=Z  «  (?)  (49). 

Similarly,  by  quadrilateral  EPSN,  it  may  be  proved  that 
Z^=Z6. 

.'.A  DEF  is  similar  to  A  PRS  (?)  (314). 

But  A  ABC  is  similar  to  A  PRS  (?)   (319). 

/.  A  ABC  is  similar  to  A  DEF  (?)  (320).  Q.E.D. 


Ex.  1.   In  the  figure  of  321,  prove  that  Z  F  =  /.  c  by  using  48. 
Ex.  2.    Draw  the  figure  for  the  theorem  of  321  if  P  is  taken  on  EF 
prolonged.     Prove  the  theorem  with  this  figure. 

Ex.  3.    Prove  the  same  theorem  if  P  is  taken  at  a  vertex. 
Ex.  4.    Prove,  if  P  is  taken  within  the  triangle  DEF. 


HOOK   ITT 


,  155 


322.   THEOREM.   Two  homologous  altitudes  of  two  similar  triangles 
are  proportional  to  any  two  homologous  sides. 

Given :  (?). 

BL          AB          AC          BC 


To  Prove: 


B'L'       A'B'       A'C'        B'C' 


B 


L' 


C1 


A  L  C       A7 

Proof :  A  ABC  is  similar  to  A  A' B'C'  (?). 

(312).   .-.AABL  is  similar  to  A  A' B'L'  (315). 


BL 


AB 


om        R 

(312)*    But' 


AB 


AC 


BC 
l^c1 


Hence, 


BL 
B'L' 


AB  _   AC  _BC^  (A       n 

A^~"A^~B'C'  ^Ax'1> 


Q.E.D. 


323.   It  is  evident  that :    In  similar  figures, 

1.  Homologous  angles  are  equal. 

2.  Homologous  sides   are   opposite   equal  angles   (in    tri- 
angles). 

Thus,   shortest  sides  are  homologous.    [Opp.  smallest  A] 
Medium  sides  are  homologous.    [Opposite  medium   A] 
Longest  sides  are  homologous.      [Opposite  largest  A] 

3.  Homologous  sides  are  proportional. 

The  antecedents  of  this  proportion  belong  to  one  of  the 
similar  figures  and  the  consequents  to  the  other. 


Ex.  1.    Prove  the  theorem  of  322  by  use  of  triangles  BLC  and  B'L'C1. 
Ex.  2.    State  all  the  instances  under  which  two  triangles  are  similar. 
Ex.  3.    In  the  figure  of  322,  if  AB  =  13,  AC  =  15,  BL  =  9,  A'C1  =  20, 
find  A'B'  and  B'L'. 


lf>0 


PLANE   GKOMKTRY 


324.  THEOREM.  If  two  parallel  lines  are  cut  by  three  or  more  trans- 
versals which  meet  at  a  point,  the  corresponding  segments  of  the  par- 
allels are  proportional. 

Given:    (?). 

To  Prove :    —  =  —  =  —  . 

CG        GH       HD 
Proof :     In  A  COG,  AE  is  ||  to 
CG  (Hyp.).     /.  A  OAE  is  simi- 
lar to  A  OCG  (?)   (316). 

Likewise,  A  OEF  is    similar 
to  A  OGH  and  A  OFB  is  simi- 
lar to  AOHD  (316). 
OE 


G 


H 


...  f-  =  ^fl(?)   (323,    3);  also   '-  L-  =  ™  (?). 
CG        OG ^  GH       OG  ^ 

*\*  lit  JtjJy          ,  Osv  T     *1  •  jLU  I    \JJi     \  J^  O         xOX 

.-.  —  =--(?).     Likewise,       —  =  (--)  =  — .   (?). 
CG       GH  GH      \OH/       HD 

Therefore,  —  =  —  =  —  (?). 
CG        GH       HD 


Q.E.D. 


325.   THEOREM.    If  three  or  more  non-parallel  transversals  intercept 
proportional  segments  on  two  parallels,  they  meet  at  a  point. 


Given:  Transversals  AB,  CD, 
EF;  parallels  AE  and  BF ;  pro- 
portion, AC  :  BD  =  CE  :  DF. 

To  Prove :  AB,  CD,  EF  meet 
at  a  point. 

Proof  :  Produce  BA  and  CD 
until  they  meet,  at  O. 

Draw  OF  cutting  AE  at  X. 


D 


Now,  AC  :  BD  =  CX  :  DF  (?)  (324). 

But  AC  :  BD  =  CE  :  DF  (Hyp.).      .'.  CX  =  CE  (?)  (300). 

Therefore,  FE  and   FX  coincide  (?)  (39). 

That  is,  FE  produced  passes  through  O.  Q.E.D. 


BOOK   III  157 

326.   THEOREM.  The  perimeters  of  two  similar  polygons  are  to  each 
other  as  any  two  homologous  sides. 


A  B  A' 

Given :  Polygon  R  whose  perimeter  =  p  and  similar 
polygon  S  whose  perimeter  =  Pr. 

To  Prove  :    P  :  P1  =  AB  :  A' B'  =  BC  :  B'c1  =  etc. 

Proof:    AB  :  A'B'  =  BC  :  Bfc'  =  CD  :  C'D'  =  etc.  (323,  3). 

.-.  AB  4-  BC  +  CD  +  etc.  :  A'B'  +  B'C*  +  C'D'  +  etc.  = 
AB  :  A'B'  =  BC  :  B'C'  =  etc.  (?)  (301). 

.-.  P  :  Pf  =  AB  :  A'B'  =  BC  :  BfCf  =  etc.  (Ax.  6).      Q.E.D. 

Ex.  1.  In  the  figure  of  324,  if  AE=EF=FB,  prove  CO  =  GH  = 
HD.  State  this  truth  in  a  theorem. 

Ex.  2.  The  median  of  a  triangle  bisects  every  line  that  is  parallel  to 
the  side  to  which  the  median  is  drawn  and  has  its  extremities  in  the 
other  sides  of  the  triangle. 

Ex.  3.  In  the  figure  of  325,  prove  that  the  line  bisecting  AC  and  BD 
will  pass  through  point  O. 

Ex.  4.   Prove  the  theorem  of  324  if  the  point  0  is  between  the  parallels. 

Ex.  5.  If  AB  and  CD  are  any  two  parallel  lines  whose  midpoints  are 
R  and  S  respectively,  prove  that  the  lines  AD,  BC,  RS  meet  in  a  point. 

Ex.  6.  Two  homologous  sides  of  two  similar  polygons  are  8  and  15. 
The  perimeter  of  the  less  polygon  is  60.  What  is  the  perimeter  of  the 
larger? 

Ex.  7.  The  perimeters  of  two  similar  polygons  are  30  and  125  re- 
spectively. If  the  shortest  side  of  the  larger  is  8$,  find  the  shortest  side 
of  the  less. 

Ex.  8.  The  sides  of  a  polygon  are  5,  6,  7.  8,  10  respectively.  Find  the 
perimeter  of  a  similar  polygon  whose  medium  side  is  17J. 


158 


PLANE   GEOMETRY 


327.  THEOREM.  If  two  polygons  are  similar,  they  may  be  decom- 
posed into  the  same  number  of  triangles  similar  each  to  each  and 
similarly  placed. 


Given  :    Similar  polygons  BE  and  B'E'. 

To  Prove  :  A  ABC  similar  to  A  A'B'C'\ 
A  ACD  similar  to  A  A'C'D'; 
A  AED  similar  to  A  A'E'D'. 

Proof  :    First.    AB  :  A'B'  =  BC  :  B'C'  (323,  3). 

Also,  Z  B  =  Z£'(323,  1). 

Therefore,  A  ABC  is  similar  to  AA'B'C'  (317). 

Second.     In  A  ABC  and  A'B'C',  -^-  =  -^-  (?)  (323,  3). 

BfC'       A'C' 

In  the  similar  polygons,  -?£  =-£®.  (?)  (323,  3). 

B  C        CD 


Consequently, 


AC 


CD 
C'D' 


(Ax.  1). 


A'C' 

In  the  polygons,  Z  BCD  =  /.B'C'D'\ 

In  the  A  ABC  and  A'B'C',  Z.BCA  =  ^  B'C'A'\   ' 
Hence,  by  subtraction,        /.ACD  =  Z.A'C'D'    (Ax.  2). 
Therefore,  A  ACD  is  similar  to  A  A' C'D'  (?)  (317). 
Third.     A  AED    is    proven    similar    to    A  A'E'D'    in    like 
manner.  Q.E.D. 


HOOK   LIT 


159 


328.   THEOREM.   If  two  polygons  are  composed  of  triangles  similar 
each  to  each  and  similarly  placed,  the  polygons  are  similar. 


K 


Given  :  A  GUI  similar  to  A  G'H'I'; 
A  GIJ  similar  to  A  G'I'J'  ; 
A  GJK  similar  to  A  GrJfKf. 

To  Prove  :    The  polygons  HK  and  HrKf  similar. 

Proof  :    First.     In  A  HGI  and  H'G'I',  Z  H  =  Z  Hf  (323,  1). 

Also  in  these  A  Z  HIG  =  Z  II'I'G'    (?)  (323,  1). 

In  A  GIJ  and  G'I'J',    Z  GIJ  =  Z  G'/'j7    (?). 

Adding,  Z  #/</  =  Z  #W     (Ax.    2). 

Likewise,  Z/JJT=  Z  I*J'K';  etc. 

That  is,  the  polygons  are  mutually  equiangular. 

Second.    In   A  GHI  and  G'H'J',  -^  =  -^  =  —  (323,  3). 

G'#'      ff'l1      G7/ 

In  A  GIJ  and  G'I'J',  -^  =  -^    (?). 
Hence,  — ^-^  =  -—  =  -^— -  (Ax.  1). 

G  U        HI        I J 


In  the  same  way,  we  may  prove 


IJ 


JK 


KG 
K'Gr 


.      GH  __    HI  _.    7J         JK  _  ,.        n 

•G7^"^?"??-/^- 

That  is,  the  homologous  sides  are  proportional. 
Therefore,  the  polygons  are  similar  (?)  (312). 


Q.E.D. 


160  PLANE   GEOMETRY 

329.  THEOREM.   If  through  a  fixed  point  within  a  circle  two  chords 
be  drawn,  the  product  of  the  segments  of  one  will  equal  the  product  of 
the  segments  of  the  other. 

Given :  Point  o  in  circle  C ; 
chords  AB  and  RS  intersecting 
at  o.  (Review  the  note,  p.  145.) 

To  Prove:   AO  •  OB  =  RO  'os. 

Proof  :    Draw  A3  arid  RB. 
In    A-4O/8    and     ROB,     /.  S  = 
(?)   (250). 

^  =  ZE  (?). 

.-.  these  A  are  similar  (?)  (314). 
Hence,    AO  :  RO  =  OS  :   OB    (?)   (323,  3). 
/.  AO  -  OB  =  RO  -  OS   (?)    (290).  Q.E.D. 

330.  THEOREM.   The  product  of  the  segments  of  any  chord  drawn 
through  a  fixed  point  within  a  circle  is  constant  for  all  chords  through 
this  point.     (See  329.) 

331.  Direct  proportion  and  reciprocal  (or  inverse)  proportion. 

Illustrations.  I.  If  a  man  earns  $4^  each  day,  in  8  days  he  will  earn 
$36.  In  12  days  he  will  earn  $5A.  Hence,  8  da.  :  12  da.  =  $36  :  $54  is 
a  proportion  in  which  the  antecedents  belong  to  the  same  condition  or 
circumstance,  and  the  consequents  belong  to  some  other  condition  or 
circumstance.  This  is  called  &  direct  proportion. 

II.  If  one  man  can  build  a  certain  wall  in  120  days,  8  men  can  build 
it  in  15  days;  or  12  men  in  10  days.  Hence,  8  men  :  12  men  =  10  da.  : 
15  da.  is  a  proportion  in  which  the  means  belong  to  the  same  condition 
or  circumstance,  and  the  extremes  belong  to  some  other  condition  or  cir- 
cumstance. This  is  called  a  reciprocal  (or  inverse)  proportion. 

Definitions.  A  direct  proportion  is  a  proportion  in  which 
the  antecedents  belong  to  the  same  circumstance  or  figure, 
and  the  consequents  belong  to  some  other  circumstance  or 
figure.  Thus  the  ordinary  proportions  derived  from  similar 
figures  are  direct  proportions.  (See  323,  3.) 

A   reciprocal   (or  inverse)   proportion   is   a  proportion   in 


BOOK    TIT 


161 


which  the  means  belong  to  the  same  circumstahce  or  figure, 
and  the  extremes  belong  to  some  other 
circumstance  or  figure. 

Thus,  in  the  adjoining  figure,  a  •  b  = 
x  •  y  (329).  .-.  a:x  =  y:b  (291). 
This  is  a  reciprocal  proportion  because 
the  means  are  parts  of  one  chord,  and  the 
extremes  are  parts  of  the  other  chord. 

332.  THEOREM.   If  through  a  fixed  point  within  a  circle  two  chords 
be  drawn,  their  four  segments  will  be  reciprocally  (or  inversely)  pro- 
portional. 

Proof  :   [Identical  with  proof  of  329;  omitting  the  last  step.] 

333.  THEOREM.    If  from  a  fixed  point  without  a  circle  a  secant  and  a 
tangent  be  drawn,  the  product  of  the  whole  secant  and  the  external  seg- 
ment will  equal  the  square  of  the  tangent. 

Given:  O  C;  secant 
PAB ;  tangent  PT. 

To  Prove : 

PB  •  'PA  =  PT2. 

Proof  :  Drawer  and  BT. 

In  A  PAT  and  PBT 
ZP=  Z  P  (Iden.). 

Z  PTA  is  measured  by 
i  arc  AT  (?).  Z  B  is 
measured  by  J  arc  AT  (?). 

Therefore,  A  PAT  is  similar  to  A  PBT  (?). 

Hence,   PB  :  PT  =  PT  :  PA  (?)  (323,  3). 

.-.   PB-PA=  PT2  (?)  (290).  Q.E.D. 

334.  THEOREM.    If  from  a  fixed  point  without  a  circle  any  secant  be 
drawn,  the  product  of  the  secant  and  its  external  segment  will  be 
constant  for  all  secants. 

Proof :    Any  secant  x  ext.  seg.  =  (tan.  )2=  constant  (333). 

BOBBINS'  PLANE  GEOM.  — 11 


162 


PLANE   GEOMETRY 


335.  THEOREM.    If  from  a  fixed  point  without  a  circle  a  secant  and 
a  tangent  be  drawn,  the  tangent  will  be  a  mean  proportional  between 
the  secant  and  its  external  segment. 

Proof  :   [Identical  with  proof  of  333;  omitting  the  last  step.] 

336.  THEOREM.    If  from  a  fixed  point  without  a  circle  two  secants 
be  drawn,  these  secants  and  their  external  segments  will  be  reciprocally 
(or  inversely)  proportional. 


B 


Proof  :  PB-PA  =  PY.PX  (?)  (334). 
/.  PB:PY=PX-.PA  (?)  (291). 


Q.E.D. 


Ex.    If  PA  =  3  in.,  and  PB  =  12  in.,  find  the  length  of  PT. 
Ex.   If  PB  =  21  in.,  PY=  15  in.,  and  PA  =  5  in.,  find  PX. 


337.  THEOREM.  In  any  triangle  the  product  of  two  sides  is  equal  to 
the  diameter  of  the  circumscribed  circle  multiplied  by  the  altitude  upon 
the  third  side. 

Given:     A    ABC:,     circum- 
scribed Oof  altitude  BK. 

To  Prove  :  a  -  c  =  d  •  h. 

Proof  :    Draw   chord    CD. 
Z  BCD  =  rt.  Z  (?)  (251). 

In  rt.  A  ABK  and  J5CZ), 
Z  A=^D  (?)  (250). 

.'.  these  A  are  similar  (?). 

.-.  c:d=h:a(?)  (323,  3). 

Consequently,  a  •  c  =  d  -  h  (?)  (290).  Q.E.D. 


BOOK   III 


163 


338.  THEOREM.  In  any  triangle  the  product  of  two  sides  is  equal 
to  the  square  of  the  bisector  of  their  included  angle,  plus  the  product 
of  the  segments  of  the  third  side  formed  by  the  bisector. 


Given :   A  ABC,   co  the 
sector  of  Z  ACB. 


bi- 


\ 


To  Prove  :  a  •  b  =  t2  +  n  -  r. 

Proof:  Circumscribe  a  O 
about  the  A  ABC. 

Produce  CO  to  meet  O  at  D ; 
draw  BD. 

In  A  AGO  and  BCD,  Z  AGO  = 
Z  BCD  (Hyp.). 

And  Z  ^  =  Z  D(?)  (250). 

.-.  A  AGO  and  BCD  are  similar  (?)  (314). 

Hence,  b  :  (t  +  x)  =  t  :  a  (?)  (323,  3). 

Therefore,  a  -  b  =  t2  +  t-  x  (?)  (290). 

CD  and  ^1B  are  chords  (Const.).      .-.  t-  x  =  n  -  r  (?)  (329). 

Consequently,   #  •  b  =  t2  +  w  •  r  (Ax.  6).  Q.E.D. 

339.   The  projection  of  a  point  upon  a  line  is  the  foot  of  the 
perpendicular  from  the  point  to  the  line 
Thus,  the  projection  of  P  is  J. 


N  M 


The  projection  of  a  definite  line  upon  an  indefinite  line  is 
the  part  of  the  indefinite  line  between  the  feet  of  the  two 
perpendiculars  to  it,  from  the  extremities  of  the  definite  line. 

The  projection  of  AB  is  CD ;   of  BS  is  RT ;   of  LM  is  NM. 


164 


PLANE   GEOMETRY 


340.  THEOREM.    If  in  a  right  triangle  a  perpendicular  be  drawn 
from  the  vertex  of  the  right  angle  upon  the  hypotenuse, 

I.    The  triangles  formed  will  be  similar  to  the  given  triangle  and 
similar  to  each  other. 

II.   The    perpendicular  will  be  a  mean  proportional  between  the 
segments  of  the  hypotenuse. 

Given:  Rt.   A  ABC-,    CP  J_  c 

to  AB  from  C.  „ 

To  Prove:   I.   A  APC,  ABC, 
and  BPC  similar. 

II.   AP  :  CP  =  CP  :  PB. 

Proof :   I.   In  rt.  A  APC  and  ABC,  Z  A  =  Z  A  (Iden.). 

/.  A  APC  is  similar  to  A  ABC  (?)   (315). 

In  rt.  A  BPC  and  ABC,  Z  B  =  /_  B  (?). 

.-.  A  BPC  is  similar  to  A  ABC  (?). 

Therefore,  A  APC,  ABC,  and  BPC  are  all  similar  (?)  (320). 

II.    In  the  A  ^4PCand  BPC,  AP  :  CP  =  CP  :  PB  (?)  (323,  3). 

Q.E.D. 

341.  THEOREM.    If  from  any  point  in  a  circumference  a  perpendicu- 
lar be  drawn  to  a  diameter,  it  will  be  a  mean  proportional  between 
the  segments  of  the  diameter. 

Given:  (?).      To  Prove :  (?). 

Proof  :  Draw  chords  AP  and  BP. 

A  APE  is  a  rt.  A  (?)  (251). 

/.  AD:PD  =  PD:  DB  (?).  Q.E.D. 

342.  THEOREM.    The  square  of  a  leg  of  a  right  triangle  is  equal  to 
the  product  of  the  hypotenuse  and  the  projection  of  this  leg  upon  the 
hypotenuse. 

Given:   Rt.  A  ABC;  AC  and 
BC  the  legs. 

To  Prove:  I.  AC?=  AB  -  AP. 

II.    2JC2  =  AB  -  BP.    A 

r\  I—  tJ 


BOOK   III 


165 


Proof:    I.  The  rt.  A  ABC  and  APC  are  similar  (?)  (340,  I). 
.  .    Ali  :  AC=  AC  :  AP  (323,  3).      .*.  JX?  =  AB  •  AP  (?). 
II.    Kt.   A  ABC  and  BCP  are  similar    (?). 
.'.  AB  :  B('=  BC:  BP  (?).      .-.!*€?=  AB  -  BP  (?).          <«>.E.D. 

Ex.  1.    If,  in  340,  .IP  =  3,  PB  =  27,  find  CP. 

Ex.  2.    If,  in  342,  AP  =  4,  PB  =  21,  find  viC  and  J5C. 

Ex.  3.    If,  in  342,  AB  =  20,  A  C  =  6,  find  JP,  £P,  CP,  and  BC. 


343.   THEOREM.    The  sum  of  the  squares  of  the  legs  of  a  right  tri- 
angle is  equal  to  the  square  of  the  hypotenuse. 

Given  :  Rt.  A  ABC.    To  Prove  :   Iff  +  7*c2  =  AB2. 
Proof  :  Draw  CP  J_  to  the  hypotenuse  AB. 
Then  AC2  =  AB  •  AP  (?)  (342). 
And  BC2  =  AB  -  BP  (?).     Adding, 


A(+  B!   =  AB  •  AP  +  AB-  BP  (Ax.  2). 

=  AB  (AP  +  BP)  =  AB-AB  =  Al?  (Ax.  4). 
That  is,  AC2  -f  BC2  =  ^1#2.  Q.E.D. 

344.   THEOREM.    The  square  of  either  leg  of  a  right  triangle  is  equal 
to  the  square  of  the  hypotenuse  minus  the  square  of  the  other  leg. 

That  is,    AC?  =  AB2  -  BC2  ;  and  BC2  =  AB2  -  A(?  (?)  (Ax.  2). 

Ex.   If  A  C  =  28  and  BC  =  45,  find  AB. 
Ex.   If  AC  =  21  and  AB  =  29,  find  BC. 


345.  THEOREM.  In  an  obtuse  triangle  the  square  of  the  side  opposite 
the  obtuse  angle  is  equal  to  the  sum  of  the  squares  of  the  other  two  sides 
plus  twice  the  product  of  one  of  these  two  sides  and  the  projection  of  the 
other  side  upon  that  one. 

Given  :  Obtuse  A  ABC ;  etc.  ^s^/'- 

Ta Prove:  e2  =  a2  -f  ft2  +  2 bp.  ^>^    /  k\ 

Proof  :    c2  =  h2  4-  (/>  +  />)2  =  ^^^            7 

W  +  p2  +  #J  +  2  bp  (?)  (343).  ^^     6            /    t>      \ 

But  7<2+jt?2=rt2(?)  (343).  A 

.\<?=a2+l2+2bp  (Ax  6).  Q.B.D. 


166 


PLANE   GEOMETRY 


346.  THEOREM.    In  any  triangle  the  square  of  the  side  opposite  an 
acute  angle  is  equal  to  the  sum  of  the  squares  of  the  other  two  sides 
minus  twice  the  product  of  one  of  these  two  sides  and  the  projection  of 
the  other  side  upon  that  one. 

Given:   (?). 

To  Prove:  #=  (?). 

Proof:  <*  =  h*+(b-py  = 

T"»        j          70      *  O  O     XO\  0 

liut,  h*-\-  jr  =  Or  (j)  ; 

.•.  <j2=a2-f  b2—  2  bp  (Ax.  6).  Q.E.D. 

NOTE.  This  theorem  is  equally  true  in  case  the  triangle  contains  an 
obtuse  angle.  Thus,  in  the  figure  of  345,  suppose  the  projection  of  AB 
is  AM=  p.  Then,  a2  =  A2  +  (p  -  6)2  =  h2  +  p*  +  b*-2bp  =  etc. 

347.  THEOREM.    I.  The  sum  of  the  squares  of  two  sides  of  a  triangle 
is  equal  to  twice  the  square  of  half  the  third  side  increased  by  twice  the 
square  of  the  median  upon  that  side. 

II.  The  difference  of  the  squares  of  two  sides  of  a  triangle  is  equal 
to  twice  the  product  of  the  third  side  by  the  projection  of  the  median 
upon  that  side. 

Given:  A  ABC',  median  =  C 

m ;  its   projection  =  p ;    and 
side  b  >  side  a. 

To  Prove: 


TT          70  O  O 

Proof:  In  A  ARC  and  BRC,  AE  =  BR  (Hyp.);  CR  is 
common;  AC  >  BC  (Hyp.).  .'.  Z  ARC  >  Z  BRC  (?)  (87). 
That  is,  Z  ARC  is  obtuse  and  Z  BRC  is  acute. 

/.  in  A  ARC,  52=  (l(?)2  +m*  +  cp  (345), 
and  in  A  BRC,  a2  =  ( J  c)2  +  m2  -  cp  (346). 

I.    Adding, 


52  +  az  =  2 
II.    Subtracting,  b2  —  a2  = 


(Ax.  2). 
(Ax.  2). 

Q.E.D. 


BOOK  III  167 

348.  If  the  vertices  of  a  triangle  are  denoted  by  A,  B,  C, 
the  lengths  of  the  sides  opposite  are  denoted  by  a,  6,  <?, 
respectively;    the   altitude   upon   these   sides  by  ha,  hb,  h^ 
respectively ;  the  bisectors  of  the  angles  by  ta,  £6,  tc,  respec- 
tively ;  the  medians  by  wa,  w6,  TWC,  respectively  ;  the  segments 
of  the  sides  formed  by  the  bisectors  of  the  opposite  angles 
by  na  and  ra,  nb  and  r6,  nc  and  rc\   and  the  projections  as 
follows :   the  projection  of   side  a  upon  side  6,  by  apb ;  of 
side  a  upon  side  (?,  by  apc;  of  side  b  upon  side  <?,  by  bpc\  etc. 

349.  Formulas.     It  is  assumed  that  a,  6,  c  are  known. 
The  following  values  of  the  various  lines  in  a  triangle  are 
obtained  by  solving  the  equations  already  derived. 

I.  Projections. 

1.  If  tc  is  obtuse,  *  =  C'-2a;-6*;  .„.  =  «-^|^  ;  etc. 

2.  If  tc  is  acute,   ,ft  =  «f +"»*  -  *  ;  tff,<t^L,  etc. 

26  2  a 

II.  Altitudes.  hb=    Va2-a/>62;  A0=  V&2-5jt>a2;  etc. 

III.  Medians.  mc  =  J  V2  (a2  +  62)  -  c2  ;  w?a  =  \  V2(62  +  c2)  -a2;  etc. 

IV.  Bisectors.   #c  =  Va6  —  ncrc* ;  £a  =  V6c  —  nara*;  £6  =  Vac  —  n6r6.* 

V.  Diameter  of  circumscribed  circle  =  ^;  =-•-!  =^' 

hb        hc         ha 

VI.  Largest  Angle. 

1.  Z  C  is  right  if  c2  =  a2  +  &2   (343). 

2.  ^  C  is  obtuse  if  c2  >  a2  +  62  (345). 

3.  Z  C  is  acute  if  c2<  a2  +  62  (346). 


Ex.  1.  If  the  sides  of  a  triangle  are  a  =  7,  &  =  10,  c  =  12,  find  the 
nature  of  Z  C. 

Ex.  2.  In  the  same  triangle  find  ma.     Find  m&.     Find  mc. 

Ex.  3.  In  the  same  triangle  find  apb.  Find  bpa.  Find  a/>c.  Find  bpc. 

Find  c/>a.  Find  cpb. 

Ex.  4.  Find  £«.     Find  hb.     Find  Ac. 

Ex.  5.  Find  the  diameter  of  the  circumscribed  circle. 

Ex.  6.  Find  na  and  ra.     Find  nb  and  r6.     Find  nc  and  TV 

Ex.  7.  Find  ta.     Find  ?5.     Find  /c. 


*  The  segments  n  and  r  can  be  found  by  308  j  m, :  rb  =  c :  a,  etc. 


168  PLANE   GEOMETRY 

CONCERNING  ORIGINALS 

350.  We  should  first  determine  from  the  nature  of  each 
numerical  exercise  upon  which  theorem  it  depends.  By 
applying  the  truth  of  that  theorem,  the  exercise  is  usually 
solved  without  difficulty. 

ORIGINAL    EXERCISES    (NUMERICAL) 

1.  The  legs  of  a  right  triangle  are  12  and  16  inches  ;  find  the  hypote- 
nuse. 

2.  The  side  of  a  square  is  6  feet;  what  is  the  diagonal? 

3.  The  base  of  an  isosceles  triangle  is  16  and  the  altitude  is  15;  find 
the  equal  sides. 

4.  The  tangent  to  a  circle  from  a  point  is  12  inches  and  the  radius 
of  the  circle  is  5  inches ;  find  the  length  of  the  line  joining  the  point  to  the 
center. 

5.  In  a  circle  whose  radius  is  13  inches,  what  is  the  length  of  a  chord 
5  inches   from   the   center? 

[Draw  chord,  distance,  and  radius  to  its  extremity.] 

6.  The  length  of  a  chord  is  2  feet  and  its  distance  from  the  center 
is  35  inches  ;  find  the  radius  of  the  circle. 

7.  The  hypotenuse  of  a  right  triangle  is  2  feet  2  inches,  and  one  leg 
is  10  inches  ;  find  the  other. 

8.  The  base  of  an  isosceles  triangle  is  90  and  the  equal  sides  are  each 
53  ;  find  the  altitude. 

9.  The  radius  of  a  circle  is  4  feet  7  inches ;  find  the  length  of  the 
tangent  drawn  from  a  point  6  feet  1  inch  from  the  center. 

10.  How  long  is  a  chord  21  yards  from  the  -center  of  a  circle  whose 
radius  is  35  yards  ? 

11.  Each  side  of  an  equilateral  triangle  is  4  feet ;  find  the  altitude. 

12.  The  altitude  of  an  equilateral  triangle  is  8  feet ;  find  the  side. 
[Let  x  =  each  side ;  \  x  =  the  base  of  each  rt.  A.] 

13.  Each  side  of  an  isosceles  right  triangle  is  a ;  find  the  hypotenuse. 

14.  If  the  length  of  the  common  chord  of  two  intersecting  circles  is 
16,  and  their  radii  are  10  and  17,  what  is  the  distance  between  their 
centers  ? 

15.  The  diagonal  of  a  rectangle  is  82  and  one  side  is  80 ;  find  the  other. 

16.  The  length  of  a  tangent  to  a  circle  whose  diameter  is  20,  from  an 
external  point,  is  26.    What  is  the  distance  from  this  point  to  the  center? 


BOOK  III  169 

17.  The  diagonal  of  a  square  is  10  ;  find  each  side. 

18.  Find   the  length  of* a  chord  '2  feet  from  the  center  of  a  circle 
whose  diameter  is  5  feet. 

19.  A  flagpole  \va.s  broken  16  feet  from  the  ground,  and  the  top  struck 
the  ground  63  feet  from  the  foot  of  the  pole.     How  long  was  the  pole? 

20.  The  top  of  a  ladder  17  feet  long  reaches  a  point  on  a  wall  15  feet 
from  the  ground.     Mow  far  is  the  lower  end  of  the  ladder  from  the  wall  ? 

21.  A  chord  2  feet  long  is  5  inches  from  the  center  of  a  circle.     How 
far  from  the  center  is  a  chord  10  inches  long?     [Find  the  radius.] 

22.  The  diameters  of  two  concentric  circles  are  1  foot  10  inches  and 
10  feet  '2  inches.     Find  the  length  of  a  chord  of  the  larger  which  is  tan- 
gent to  the  less. 

23.  The  lower  ends  of  a  post  and  a  flagpole  are  42  feet  apart ;  the 
post  is  8  feet  high  and  the  pole,  48  feet.     How  far  is  it  from  the  top  of 
one  to  the  top  of  the  other? 

24.  The  radii  of  two  circles  are  8  inches  and  17  inches,  and  their  cen- 
ters are  41  inches  apart.     Find  the  lengths  of  their  common  external 
tangents;   of  their  common  internal  tangents. 

25.  A  ladder  65  feet  long  stands  in  a  street ;  if  it  inclines  toward  one 
side,  it  will  touch  a  house  at  a  point  16  feet  above  the  pavement;  if  to 
the  other  side,  it  will  touch  a  house  at  a  point  56  feet  above  the  pave- 
ment.    How  wide  is  the  street  ? 

26.  Two  parallel  chords  of  a  circle  are  4  feet,  and  40  inches  long,  re- 
spectively, and  the  distance  between  them  is  22  inches.     Find  the  radius 
of  the  circle. 

[Draw  the  radii  to  ends  of  chords;    these  =  hypotenuses  =  72;   the 
distances  from  the  center  =  x  and  22  —  x.~\ 

27.  The  legs  of  an  isosceles  trapezoid  are  each  2  feet  1  inch  long,  and 
one  of  the  bases  is  3  feet  4  inches  longer  than  the  other.     Find   the 
altitude. 

28.  One  of  the  non-parallel  sides  of  a  trapezoid  is  perpendicular  to 
both  bases,  and  is  63  feet  long ;    the  bases  are  41  feet  and  25  feet  long. 
Find  the  length  of  the  remaining  side. 

29.  If  a  =  10,  h  =    6,  find  />,  c,  //,  ft. 

30.  If  h  =    8,  />'  =    4,  find  ft,  c,  p,  a. 

31.  If  a  =  10,  p'  =  15,  find  c,  p,  h,  ft. 

32.  If  «  =    0,  ft   =  12,  find  c,  />,  />',  h. 

33.  If  p  =    3,  p'  =  12,  find  a,  /*,  ft. 


170  PLANE   GEOMETRY 

34.  The  line  joining  the  midpoint  of  a  chord  to  the  midpoint  of  its 
arc  is  5  inches.     If  the  chord  is  2  feet  long,  what  is  the  diameter? 

35.  If  the  chord  of  an  arc  is  60  and  the  chord  of  its  half  is  34,  what  is 
the  diameter  ? 

36.  The  line  joining  the  midpoint  of  a  chord  to  the  midpoint  of  its 
arc  is  6  inches.     The  chord  of  half  this  arc  is  18  inches.     Find  the 
diameter.     Find  the  length  of  the  original  chord. 

37.  To  a  circle  whose  radius  is  10  inches,  two  tangents  are  drawn 
from  a  point,  each  2  feet  long.     Find  the  length  of  the  chord  joining 
their  points  of  contact. 

38.  The  sides  of  a  triangle  are  6,  9,  11.     Find  the  segments  of  the 
shortest  side  made  by  the  bisector  of  the  opposite  angle. 

39.  Find  the  segments  of  the  longest  side  made  by  the  bisector  of  the 
largest  angle  in  No.  38. 

40.  The  sides  of  a  triangle  are  5,  9,  12.     Find  the  segments  of  the 
shortest  side  made  by  the  bisector  of  the  opposite  exterior  angle.     Also 
of  the  medium  side  made  by  the  bisector  of  its  opposite  exterior  angle. 

41.  In  the  figure  of  306,  if  AC  =  3,  CE  =  5,  EG  =  8,  BD  =  4;  find 
DF  and  FH. 

42.  If  the  sides  of  a  triangle  are  6,  8, 12  and  the  shortest  side  of  a 
similar  triangle  is  15,  find  its  other  sides. 

43.  If  the  homologous  altitudes  of  two  similar  triangles  are  9  and  15 
and  the  base  of  the  former  is  21,  what  is  the  base  of  the  latter? 

44.  In  the  figure  of  324,  AE  =  4,  EF=Q,  FB=Q,  £#=15.     Find 
CG  and  CD. 

45.  The  sides  of  a  pentagon  are  5,  6,  8,  9,  18,  and  the  longest  side  of 
a  similar  pentagon  is  78.     Find  the  other  sides. 

46.  A  pair  of  homologous  sides  of  two  similar  polygons  are  9  and  16. 
If  the  perimeter  of  the  first  is  117,  what  is  the  perimeter  of  the  second  ? 

47.  The  perimeters  of  two  similar  polygons  are  72  and  120.     The 
shortest  side  of  the  former  is  4,  what  is  the  shortest  side  of  the  latter  ? 

48.  Two  similar  triangles  have  homologous  bases  20  and  48.     If  the 
altitude  of  the  latter  is  36,  find  the  altitude  of  the  former. 

49.  The  segments  of  a  chord,  made  by  a  second  chord,  are  4  and  27. 
One  segment  of  the  second  chord  is  6,  find  the  other. 

50.  One  of  two  intersecting  chords  is  19  in.  long  and  the  segments  of 
the  other  are  5  in.  and  12  in.     Find  the  segments  of  the  first  chord. 

51.  Two  secants  are  drawn  to  a  circle  from  a  point;  their  lengths  are 


BOOK   Til  171 

15  inches  and  10|  inches.     The  external  segment  of  the  latter  is  10;  find 
the  external  segment  of  the  former. 

52.  The  tangent  to  a  circle  is  1  foot  long  and  the  secant  from  the 
same  point  is  1  foot  6  inches.     Find  the  chord  part  of  the  secant. 

53.  The  internal  segment  of  a  secant  25  inches  long  is  16  inches. 
Find  the  tangent  from  the  same  point  to  the  same  circle. 

54.  Two   secants  to  a  circle  from  a  point   are   1£  feet  and  2  feet 
long ;  the  tangent  from  the  same  point  is  12  inches.     Find  the  external 
segments  of  the  two  secants. 

55.  The  sides  of  a  triangle  are  5,  6,  8.     Is  the  angle  opposite  8  right, 
acute,  or  obtuse  ?     Same  for  the  triangle  8,  7,  4  ? 

56.  The  sides  of  a  triangle  are  8,  9,  12.     Is  the  largest  angle  right, 
acute,  or  obtuse  ?     Same  for  the  triangle  13,  7,  11  ? 

57.  The  sides  of  a  triangle  are  z,  y,  z.     If  z  is  the  greatest  side,  when 
will  the  angle  opposite  be  right?    Obtuse?     Acute? 

58.  The  sides  of  a  triangle  are  6,  8,  9.     Find  the  length  of  the  projec- 
tion of  side  6  upon  side  8 ;  of  side  8  upon  side  9 ;  of  side  9  upon  side  6. 

59.  The  sides  of  a  triangle  are  5,  6,  9.     Find  the  length  of  the  pro- 
jection of  side  6  upon  side  5 ;  of  side  9  upon  side  6. 

60.  Find  the  three  altitudes  in  triangle  9,  10,  17. 

61.  Find  the  three  altitudes  in  triangle  11,  13,  20. 

62.  Find  the  diameter  of  circumscribed  circle  about  triangle  17,  25,  26. 

63.  Find  the  length  of  the  bisector  of  the  least  angle  of  triangle 
7,  8,  20.     Also  of  the  largest  angle. 

64.  Find  the  length  of  the  bisector  of  the  largest  angle  of  triangle 
12,  32,  33 ;  also  of  the  other  angles. 

65.  Find  the  three  medians  in  triangle  4,  7,  9. 

66.  Find  the  product  of  the  segments  of  every  chord  drawn  through 
a  point  4  units  from  the  center  of  a  circle  whose  radius  is  10  units. 

67.  The  bases  of  a  trapezoid  are  12  and  20,  the  altitude  is  8;  the 
other  sides  are  produced  to  meet.     Find  the  altitude  of  the  larger  tri- 
angle formed. 

68.  The  shadow  of  a  yardstick  perpendicular  to  the  ground  is  4£  feet. 
Find  the  height  of  a  tree  whose  shadow  at  the  same  time  is  100  yards. 

69.  There  are  two  belt-wheels  3  feet  8  inches  and  1  foot  2  inches  in 
diameter,  respectively.     Their  centers  are  9  feet  5  inches  apart.     Find 
the  length  of  the  belt  suspended  between  the  wheels  if  the  belt  does  not 
cross  itself.     Also  the  length  of  the  belt  if  it  does  cross. 


172  PLANE  GEOMETRY 


SUMMARY 

351.  Triangles  are  proven  similar  by  showing  that  they  have : 

(1)  Two  angles  of  one  equal  to  two  angles  of  the  other. 

(2)  An  acute  angle  of  one  equal  to  an  acute  angle  of  the  other.     [In 
right  triangles.] 

(3)  Homologous  sides  proportional. 

(4)  An  angle  of  one  equal  to  an  angle  of  the  other  and  the  including 
sides  proportional. 

(5)  Their  sides  respectively  parallel  or  perpendicular. 

352.  Four  lines  are  proven  proportional  by  showing  that  they  are  : 

(1)  Homologous  sides  of  similar  triangles. 

(2)  Homologous  sides  of  similar  polygons. 

(3)  Homologous  lines  of  similar  figures. 

353.  The  product  of  two  lines  is  proven  equal  to  the  product  of  two 
other  lines,  by  proving  these  four  lines  proportional  and    making   the 
product  of  the  extremes  equal  to  the  product  of  the  means. 

354.  One  line  is  proven  a  mean  proportional  between  two  others  by 
proving   that   two   triangles   which   contain   this    line  in   common   are 
similar,  and  obtaining  the  required  proportion  from  their  sides. 

355.  In  cases  dealing  with  the  square  of  a  line,  one  uses : 

(1)  Similar  triangles  having  this  line  in  common,  or, 

(2)  A  right  triangle  containing  this  line  as  a  part. 

ORIGINAL   EXERCISES   (THEOREMS) 

1.  If  two  transversals  intersect  between  twro  parallels,  the  triangles 
formed  are  similar.     [Use  351  (1).] 

2.  Two  isosceles  triangles  are  similar  if  a  base  angle  of  one  is  equal 
to  a  base  angle  of  the  other. 

3.  Two  isosceles  triangles  are  similar  if  the  vertex-angle  of  one  is 
equal  to  the  vertex -angle  of  the  other. 

4.  The  line  joining  the  midpoints  of  two  sides  of  a  triangle  forms  a 
triangle  similar  to  the  original  triangle. 

5.  The  diagonals  of  a  trapezoid  form,  with  the  parallel  sides,  two 
similar  triangles. 


BOOK   III 


173 


6.  Two  circles  are  tangent  externally  at  P;  through  P  three  lines  are 
drawn,  meeting  one  circumference  in  .4,  B,  C, 

and  the    other   in  A',  B',  C".     The    triangles 

ABC  and  A'B'C'  are  similar.  .        _     

7.  Prove  the  same  theorem  if  the  circles 
are  tangent  internally. 

8.  If  two  circles  are  tangent  externally  at 

P,  and  BB',  CC'  be  drawn  through  P,  terminating  in  the  circumferences, 
the  triangles  PBC  and  PB'C'  will  be  similar. 
[Draw  the  common  tangent  at  P.] 

9.  Prove  the  same  theorem  if  the  circles  are  tangent  internally. 

10.  If  AD  and  BE  are  two  altitudes  of  triangle 
ABC,  the  triangles  ACD  and  BCE  are  similar. 

11.  Two  altitudes  of  a  triangle  are  reciprocally 
proportional  to  the  bases  to  which  they  are  drawn. 

To  Prove  :    AD  :  BE  =  A  C  :  BC. 

12.  The  four  segments  of  the  diagonals  of  a  trapezoid  are  proportional 

13.  If  at  the  extremities  of  a  right  triangle  perpen- 
diculars be  erected  meeting  the  legs  produced,  the  new 
triangles  formed  will  be  similar. 

14.  In  the  figure  of  Xo.  13,  prove  : 

(1)  Triangle   ABC  similar  to  each  of  the  triangles 
A  CE  and  BCD. 

(2)  Triangle  ABE  similar  to  triangle  ABD. 

(3)  Triangle  ACE  similar  to  triangle  ABD. 

(4)  Triangle  BCD  similar  to  triangle  ABE. 

(5)  Triangles  ABC,  ABD,  ABE  similar. 

15.  If  AD  and  BE  are  two  altitudes  of  triangle  ABC  (fig.  of  No.  11), 
meeting  at  O,  the  triangles  BOD  and  A  OE  are  similar. 

16.  Triangles  CED  and  ABC   (fig.  of  No.  11)  are 
similar. 

[First  show  A  CA  D  and  CEB  similar . 

/.  CA  :  CB  =  CD  :  CE  (?).     Then  use  351  (4).] 

17.  Triangle  ABC  is  inscribed  in  a  circle  and  AP  is 
drawn  to  P,  the  midpoint  of  arc  BC,  meeting  chord  CB 
at  D.     The  triangles  A  BD  and  A  CP  are  similar. 


B 


174 


PLANE   GEOMETRY 


18.  Two   homologous   medians  in  two  similar  triangles  are  in  the 
same  ratio  as  any  two  homologous  sides. 

[Prove  a  pair  of  the  new  triangles  formed,  similar,  by  351  (4).] 

19.  Two  homologous  bisectors  in  two  similar  triangles  are  in  the  same 
ratio  as  any  two  homologous  sides. 

20.  The  radii  of  circles  inscribed  in  two  similar  triangles  are  in  the 
same  ratio  as  any  two  homologous  sides. 

[Bisect  two  pairs  of  homol.  A ;  draw  the  altitudes  of  these  new  & ;  etc.] 

21.  The  radii  of  circles  circumscribed  about  two  similar  triangles  are 
in  the  same  ratio  as  any  two  homologous  sides. 

[Erect  J_  bisectors ;  draw  radius  in  each  O.] 

22.  In  any  right  triangle  the  product  of  the  hypotenuse  and  the  al- 
titude upon  it  is  equal  to  the  product  of  the  legs. 

23.  If  two  circles  intersect  at  A  and  B  and  A  C 
and  AD  be  drawn  each  a  tangent  to  one  circle  and 
a  chord  of  the  other,  the  common  chord  AB  will  be 
a  mean  proportional  between  BC  and  ED. 

24.  If  two  circles  are  tangent  externally,  the  chords  formed  by  a 
straight  line  drawn  through  their  point  of  contact  have  the  same  ratio 
as  the  diameters  of  the  circles. 

[Draw  com.  tang,  at  point  of  contact ;  draw  di- 
ameters from  point  of  contact ;  prove  A  sim. ;  etc.] 

25.  If  AB  is  a  diameter  and  BC  a  tangent,  and 
AC  meets  the  circumference  at  D,  the  diameter  is 
a  mean  proportional  between  A  C  and  AD. 

[Draw  BD.     Prove  A  containing   AB  similar.] 

26.  If  a  tangent  be  drawn  from  one  extremity  of  a 
diameter,  meeting  secants  from  the  other  extremity, 
these  secants  and  their  internal  segments  will  be  recip- 
rocally proportional. 

To  Prove:    AC:  AD  =  AS:  AR. 
Proof:    Draw  RS.     In  A  ARS  and  ACD,  ^.A  = 
Z.  A  and  Z.  ARS  =  Z  D.     (Explain.)     Etc. 

27.  If  AB  is  a  chord  and  CE,  another  chord,  drawn 
from  C,  the  midpoint  of  arc  AB,  meeting  chord  AB  at 
D,  A  C  is  a  mean  proportional  between  CD  and  CE. 

Prove  the  above  theorem  and  deduce  that,  CE  •  CD 
is  constant  for  all  positions  of  the  point  E  on  arc  A  EB. 


BOOK  III 


175 


28.  If  chord  A  D  be  drawn  from  vertex  A  of  inscribed 
isosceles  triangle  ABC,  cutting  EC  at  E,  AB  will  be 
a  mean  proportional  between  AD  and  AE. 

Prove  the  above  theorem  and  deduce  that,  AD  •  AE 
is  constant  for  all  positions  of  the  point  D  on  arc  BDC. 

29.  If  a  square  be  inscribed  in  a  right  triangle  so 
that  one  vertex  is  on  each  leg  of  the  triangle  and 

the  other  two  vertices  on  the  hypotenuse,  the  side 
of  the  square  will  be  a  mean  proportional  between 
the  other  segments  of  the  hypotenuse. 

To    Prove:   A  D :  DE  =  DE :  EB.     First  prove 
&ADG  and  BEF  similar. 

30.  If  the  sides  of  two  triangles  are  respectively  parallel,  the  lines 
joining  homologous  vertices  meet  in  a  point.     (These  lines  to  be  pro- 
duced if  necessary.) 


31.  In  each,  of  the  following  triangles,  is  the  greatest  angle  right, 
acute,  or  obtuse,  7,  24,  25  ?    13,  10,  8  ?     19,  13,  23  ? 

32.  Prove  theorem  of  329  by  drawing  two  other  auxiliary  chords. 

33.  Prove  theorem  of  325  if  point  O  is  between  the  parallels. 

34.  Prove  theorem  of  336  by  drawing  A  Y  and  BX. 

35.  In  any  triangle  the  difference  of  the 
squares  of  two  sides  is  equal  to  the  difference  of 
the  squares  of  their  projections  on  the  third  side. 

[AB2  =  (?) ;  BC2  =  (?) .     Subtract,  etc.] 

36.  If  the  altitudes  of  triangle  ABC  meet  at 
0,  AB2-  AC2  =  B02-  CO2. 

[Consult  No.  35  and  substitute.] 

37.  The  square  of  the  altitude  of  an  equilateral  triangle  is  three 
fourths  the  square  of  a  side.     [Let  side  =  a,  etc.] 

38.  If  one  leg  of  a  right  triangle  is  double 
the  other,  its  projection  upon  the  hypotenuse 
is  four  times  the  projection  of  the  other. 

Proof:  (2a)2  =  c/>;  a?  =  cp'  (?). 


D 


/.  p  =    £- ;  p>  =       (Ax.  3). 


176 


PLANE   GEOMETRY 


39.  If  the  bisector  of  an  angle  of  a  triangle  bisects  the  opposite  side, 
the  triangle  is  isosceles. 

40.  The  tangents  to  two  intersecting  circles  from  any  point  in  their 
common  chord  produced  are  equal.     [Use  333.] 

41.  If  two  circles  intersect,  their  common   chord,  produced,  bisects 
their  common  tangents.     [Use  333.] 

42.  If  A B  and  A  C  are  tangents  to  a  circle 
from  A ;    CD   is   perpendicular   to   diameter 
BOX  from  C;  then  AB  .  CD  =  BD  .  BO. 

[Use  351  (5).] 

43.  If  the  altitude  of  an  equilateral   tri- 
angle is  h,  find  the  side.    [Denote  the  side  by  x  and  half  the  base  by  £  x.~\ 

44.  If  one  side  of  a  triangle  be  divided  by  a  point  into  segments 
which  are  proportional  to  the  other  sides,  a  line  from  this  point  to  the 
opposite  angle  will  bisect  that  angle.     [Converse  of  308.] 

To  Prove :  /.  n  =Z  m  in  fig.  of  308. 

Proof:  Produce  CB  to  P,  making  BP  =  AB;  draw  AP]  etc. 

45.  State  and  prove  the  converse  of  310. 

46.  Two  rhombuses   are  similar  if  an  angle  of  one  is  equal  to  an 
angle  of  the  other. 

47.  If  two  circles  are  tangent  internally  and  any  two  chords  of  the 
greater  be  drawn  from  their  point  of  contact,  they  will  be  divided  propor- 
tionally by  the  circumference  of  the  less. 

[Draw  diameter  to  point  of  contact  and  prove  the  right  &  similar.] 

48.  The  non-parallel  sides  of  a  trapezoid  and  the  line  joining  the  mid- 
points of  the  bases,  if  produced,  meet  at  a  point.     [Use  method  of  325.] 

49.  The  diagonals  of  a  trapezoid  and  the  line  joining  the  midpoint 
of  the  bases  meet  at  a  point. 

50.  If  one  chord  bisects  another,  either  segment  of  the  latter  is  a  mean 
proportional  between  the  segments  of  the  other. 

51.  Two  parallelograms  are  similar  if  they  have  an  angle  of  the  one 
equal  to  an  angle  of  the  other  and  the  including  sides  proportional. 

52.  Two  rectangles  are  similar  if  two  adjoining  pairs  of  homologous 
sides  are  proportional. 

53.  If  two  circles   are  tangent  externally, 
the  common  exterior  tangent  is  a  mean  pro- 
portional between  the  diameters. 

[Draw  chords  PA,  PC,  PB,  PD.  Prove,  first, 
APD  and  BPC  straight  lines.  Second,  &ABC 
and  ABD,  similar.] 


HOOK  ill  177 

54.  In  any  rhombus  the  sum  of  the  squares  of  the  diagonals  is  equal 
to  the  square  of  half  the  perimeter. 

55.  If  in  an  angle  a  series  of  parallel  lines  be  drawn  having  their  ends 
in  the  sides  of  the  angle,  their  midpoints  will  lie  in  one  straight  line. 

56.  If  ABC  is  an  isosceles  triangle  and  BX  is  the  altitude  upon  AC 
(one  of  the  legs),  BC2  =  2  A  C  -  CX.     [Use  346.] 

57.  In  an  isosceles  triangle  the  square  of  one  leg  is 
equal  to  the  square  of  the  line  drawn  from  the  vertex 
to  any  point  of  the  base,  plus  the  product  of  the  seg- 
ments of  the  base. 

Proof  :  Circumscribe  a  O ;  use  method  of  338. 

58.  If  a  line  be  dra*wn  in  a  trapezoid  parallel  to  the  bases,  the  seg- 
ments  between    the   diagonals    and   the   non-  _  c 
parallel  sides  will  be  equal. 

Proof  :    &  AHI  and  ABC  are  similar  (?) ; 

&DJ#  and  DOB  also  (?).  /.^L?  =  ^(?). 

AB      BC  A  -D 

T\V        V  J  A  ff         DJT 

/c  =  Jd (?>    But' if  =  l£  (?)'  (Use  Ax* 1} ;  etc> 

59.  A  line  through  the  point  of  intersection  of  the  diagonals  of  a 
trapezoid,  and  parallel  to  the  bases,  is  bisected  by  that  point. 

60.  If  M  is  the  midpoint  of  hypotenuse  AB  of  right  triangle  ABC, 
AH?  +  EC2-  +  AC2  =  8  CJ/2. 

61.  The  squares  of  the  legs  of  a  right  triangle  have  the  same  ratio  as 
their  projections  upon  the  hypotenuse. 

62.  If  the  diagonals  of  a  quadrilateral  are  perpendicular  to  each  other, 
the  sum  of  the  squares  of  one  pair  of  opposite  sides  is  equal  to  the  sum  of 
the  squares  of  the  other  pair. 

63.  The  sum  of  the  squares  of  the  four  sides  of  a  parallelogram  is 
equal  to  the  sum  of  the  squares  of  the  diagonals.     [Use  347,  I.] 

64.  If  DE  be  drawn  parallel  to  the  hypotenuse 
A B  of  right  triangle  ABC,  meeting  AC  at  D  and 
CB  at  E,  AE2  +  ~BD2  =  AB2  +  DE2. 

[Use  4  rt.  A  having  vertex  C.] 

65.  If  between  two   parallel  tangents  a  third 
tangent  be  drawn,  the  radius  will  be  a  mean  propor- 
tional between  the  segments  of  the  third  tangent. 

To  Prove  :  BP  :  OP  =  OP  :  PD.     Proof :  A  BOD  is  a  rt.  A  (?).    Etc. 

BOBBINS*    PLANE    GEOM.  12 


178 


PLANE   GEOMETRY 


66.  If  A  BCD  is  a  parallelogram,  BD  a  diagonal,  A  G  any  line  from  A 
meeting  BD  at  E,  CD  at  F,  and  BC  (pro- 
duced) at  G,  AE  is  a  mean  proportional  be- 
tween EF  and  EG. 

Proof  :  &  A  BE  and  £Z)F  are  similar  (?)  ; 
also  &ADE  and  BEG  (?).    Obtain  two  ratios     j 
-BE\ED  and  then  apply  Ax.  1.  A 

67.  An  interior  common  tangent  of  two  circles  divides  the  line  join- 
ing their  centers  into  segments  proportional  to  the  radii. 

68.  An  exterior  common  tangent  of  two  circles  divides  the  line  join- 
ing their  centers  (externally)  into  segments  proportional  to  the  radii. 

69.  The  common  internal  tangents  of  two  circles  and 
the  common  external  tangents  meet  on  the  line  deter- 
mined by  the  centers  of  the  circles. 

70.  If  from  the  midpoint  P,  of  an  arc  subtended  by 
a  given  chord,  chords  be  drawn  cutting  the  given  chord, 
the  product  of  each  whole  chord  from  P  and  its  segment 
adjacent  to  P  will  be  constant. 

Proof:    Take  two  such  chords,  PA  and  PC;  draw  diameter  PX ;  etc. 
Rt.  &  PST  and  PCX  are  similar.  (Explain.) 

71.  If  from  any  point  within  a  triangle  ABC,  perpendiculars  to  the 
sides  be  drawn  —  OR  to  AB,  OS  to  BC,  OT  to  A  C,  AR2  +  BS2  +  ~CT2 
=  BR2  +  CS2  +  AT2.    [Draw  A O,  BO,  CO.] 

72.  If  two  chords  intersect  within  a  circle  and  at 
right  angles,  the  sum  of  the  squares  of  their  four  seg- 
ments equals  the  square  of  the  diameter. 

To   Prove:    AP2  +  BP2  +  CP2  +  DP2  =  AR2- 
Proof:  Draw  BC,  AD,  RD.     Ch.  BR  is  JL  to  AB  (?). 
/.  CD  is    ||  to  BR    (?).     .'.  arc  BC  —  arc  RD  (?). 
Hence,  ch.  BC  =  ch.   RD  (?).     Now,   RD2  =  BC2  =  BP2  +  CP2  (?). 
;fD2=  etc.   (?).    Finally,  AR2  =  AD2  +  RD2  =  etc.  (?). 

73.  The  perpendicular  from  any  point  of  an  arc  upon  its  chord  is  a 
mean    proportional    between    the    perpendiculars 

from  the  same  point  to  the  tangents  at  the  ends 
of  the  chord. 

To  Prove :  PR  :  PT  =  PT :  PS.  Proof :  Prove 
A  ARP  and  BTP  are  sim.,  also  A  APT  and 
PBS~(1).  Thus,  get  two  ratios  each  =  PA  :  PB. 


BOOK  III 


179 


74.  If  lines  be  drawn  from  any  point  in  a  circumfer- 
ence to  the  four  vertices  of  an  inscribed  square,  the  sum 
of  the  squares  of  these  four  lines  will  be  equal  to  twice 
the  square  of  the  diameter. 

Proof:  A  A  PC,  DPB,  are  rt.  A;  etc. 

75.  If  lines  be  drawn  from  any  external  point  to  the 
vertices  of  a  rectangle    ABCD,  the   sum  of  the 
squares  of  two  of  them  which  are  drawn  to  a  pair 

of  opposite  vertices  will  be  equal  to  the  sum  of  the 
squares  of  the  other  two. 

To  Prove  :  PA'2  +  PC2  =  PB*  +  PD*. 

Proof :  Draw  PEF  ±  to  the  base,  etc. 

76.  Is  the  theorem  of  No.  75  true  if  the  point 
is  taken  within  the  rectangle  ? 

77.  If  each  of  three  circles  intersects  the  other 
two,  the  three  common  chords  meet  in  a  point. 

Given  :  (?).  To  Prove :  AB,  LM,  RS  meet  at  0. 
Proof :  Suppose  A  B  and  LM  meet  at  0.  Draw  RO 
and  produce  it  to  meet  the  ©  at  X  and  X'.  Prove 
OX  =  OX'  (by  329).  .-.  X,  X',  S  are  coincident. 

78.  In  an  inscribed  quadrilateral  the  sum  of  the 
products  of  the  two  pairs  of  opposite  sides  is  equal  to 
the  product  of  the  diagonals. 

Proof:  Draw  DX  making  Z  CDX  =  Z  ADB', 
&  ADB  and  CDX  are  sim.  (?)  ;  also  A  BCD  and 
ADX  (?).  Hence,  AB  -  DC  =  DB  •  XC  (?),  and 
AD  •  BC  =  DB  •  AX  (?).  Adding  ;  etc. 

79.  If  AB  is  a  diameter,  BC  and  AD  tangents,  meeting  chords  AF 
and  BF  (produced)  at  C  and  D  respectively,  AB  is  a  mean  proportional 
between  the  tangents  BC  and  AD. 

80.  If  ABO  is  isosceles,  AB  =  CO,  and  AO  :  CO  -  CO  :  AC,  prove: 

(1)  AB  =  BC  =  CO ;  (2)  Z  ABO  =  2  Z  0 ;  (3)  Z  0  =  36°. 
Proofs:    (1)    AO:AB  =  AB:AC(Ax.  6).     /.  A  ABC 

is  sim.  to  A  A  BO  (?)  (317).    /.  A  A  BC  is  isosceles  (?) .     Etc. 

(2)  Z   ABC  =  Z   0    (?),  and   Z    CBO  =  Z   0    (?). 
/.  Z  ABO  =  2  Z  0  (Ax.  2). 

(3)  A  0  =  I  the  sum  of  A  of  A  A  OB  (?).     Etc. 

81.  If  from  a  point  A  on  the  circumference  of  a  circle  two  chords  be 
drawn  and  a  line  parallel  to  the  tangent  at  A  meet  them,  the  chords 
and  their  segments  nearer  to  A  will  be  inversely  proportional. 


180  PLANE   GEOMETRY 

CONSTRUCTIONS 

356.  PROBLEM.    To  find  a  fourth  proportional  to  three  given  lines. 

Given  :   Three  lines,  a,  b,  c.    Q ...--B 

Required :  To  find  a  fourth    bn  "" ^    ...-•%"' 

proportional  to  a,  b,  c.  ...-••'' 

R  ••'"'" 

Construction  :  Take  two  in-  ^    ...•••( 

definite    lines,    AB    and    AC,       .,.;;:'.'1<? JE \ c 

meeting  at  A.     On  AB  take 

AE  =  a,  BV=b.     On  AC  take  AS  =  c.     Draw  BS. 

From  V  draw  VW  II  to  BS,  meeting  AC  at  W. 

Statement :   SW  is  the  fourth  proportional  required.    Q.E.F. 

Proof:   In  A  AVW,  BS  is  II  to  VW  (Const.). 

.-.  a  :  b  =  c  :  SW  (?)  (302).  Q.E.D. 

357.  PROBLEM.    To  find  a  third  proportional  to  two  given  lines. 
Given:   (?).  B 
Required:  (?).                                                 v/' 
Construction :                                                £••*'     \ 

Like  that  for  356.  o/\ 

Statement:   (?).     Proof:  (?).     *""""£ s — ~ w c 

358.  PROBLEM.    To  divide  a  given  line  into  segments  proportional 
to  any  number  of  given  lines.        ^ j__        H      G  B 

Given  :  AB ;   a,  b,  c,  d.  "\/ 

Required :   To  divide  AB 

into   parts   which   shall   be     a b'" 

proportional  to  a,  b,  c,  d.  h 

Construction  :    Draw  AX,  F 

an  indefinite    line,    oblique  '"'X 

to  AB  from  A.  On  AX  take  AC  =  a,  CD  =  b,  DE  =  c,  EF  =  d. 
Draw  FB.  Then  draw,  through  E,  D,  and  C,  lines  II  to  BF, 
as  EG,  DH,  and  Cl. 

Statement:   Ai,  IH,  HG,  GB  are  the  required  parts.     Q.E.F. 

Proof:    Ai:  a  =  in :  b  =  HG  :  c  =  GB  :  d  (?)  (306).      Q.ED. 


BOOK    III 


181 


359.    PROBLEM.    To  construct  a  triangle  similar  to  a  given  triangle 
and  having  a  given  side  homologous  to  a  side  of  the  given  triangle. 

C  V,  ,x 

""••J/ 


A  B    R 

Given  :  A  ABC  and  RS  homologous  to  AB. 
Required :  To  construct  a  A  on  RS  similar  to  A  ABC. 
Construction :  At   R   construct  Z  SRX  =  Z  A ;    at    s   con- 
struct Z  R$Y  =  /_  B,  the  sides  of  these  angles  meeting  at  T. 
Statement:   (?).     Proof:   (?)  (314). 

360.   PROBLEM.    To  construct  a  polygon  similar  to  a  given  polygon 
and  having  a  given  side  homologous  to  a  side  of  the  given  polygon. 
D 


D' 


A  B  A' 

Given  :   Polygon  EB ;  line  A'B'  homologous  to  AB. 
Required :  To  construct  a  polygon  upon  AfBr,  similar  to 
polygon  EB. 

Construction  :   From  A  draw  diagonals  AC  and  AD. 
On  A'B'  construct  A  ArBfCf  similar  to  A  ABC  (by  359). 
On  A'c'  construct  A  A'c'D1  similar  to  A  A  CD.     Etc. 

Statement:  (?). 

Proof:   (?)  (328). 


182 


PLANE  GEOMETRY 


....y 


361.  PROBLEM.    To  find  the  mean  proportional  between  two  given 
lines.  rv    

Given  :  Lines  a  and  b. 

Required :  To  find  the 
mean  proportional  between 
them. 

Construction  :  On  an  in- 
definite line,  AX,  take  AB 

=  a  and  BC  =  b.  Using  o,  the  midpoint  of  AC,  as  center,  and 
AO  as  radius,  describe  the  semicircumference,  ADC.  At  B 
erect  BD  _L  to  AC,  meeting  the  arc  at  D.  Draw  AD  and  CD. 

Statement:  BD  is  the  mean  proportional  required.     Q.E.F. 

Proof  :    a  :  BD  =  BD  :  b  (?)  (341).  Q.E.D. 

362.  A  line  is  divided  into  extreme  and  mean  ratio  if  one 
segment  is  a  mean  proportional  between  the  whole  line  and 
the  other  segment.      In  other  words,  if  a  line  is  to  one  of  its 
parts,  as  that  part  is  to  the  other  part,  the  line  is  divided 
into  extreme  and  mean  ratio. 

363.  PROBLEM.    To  divide  a  line  into  extreme  and  mean  ratio. 
Given:  Line  AB  =  a. 

R 

Required  :  To  divide  ..••" ] •••.. 

AB  into  extreme  and 
mean  ratio,  that  is,  so 
that  AB:  AF=AF:FB. 

Construction :  At  B 
erect  BR,  J_  to  AB  and 
=  AB.  Using  C,  the 
midpoint  of  BR,  as  cen- 
ter, and  CB  as  radius,  *£. 
describe  a  O.  Draw 
AC  meeting  O  at  D  and  E.  On  AB  take  AF  =  AD  ;  let  AF=  x. 

Statement :  F  divides  AB  so  that  AB  :  AF  =  AF:  FB.   Q.E.F. 


\E 


a- x 


BOOK  III 


183 


Proof:  DE=a  (?)  (203).     /.  AE=  a  +  x  (Ax.  4). 
AB  is  tangent  to  the  O  (?)  (215). 

?)  (333). 
x.  6). 
That  is,  3?  =  a2  —  ax,  or  x2  =  a  (a  —  x). 

.-.  a:x  =  x:  a-x  (?)  (291). 
That  is,   AB  :  AF  =  AF:  FB  (Ax.  6).  Q.E.D. 

364.   PROBLEM.    To  divide  a  line  externally  into  extreme  and  mean 
ratio. 

Given:  (?). 

Required:  (?). 

Construction:  The  same  as  in 
363,  except  AFr  is  taken  on  BA 
produced,  =  AE. 


Q.E.F. 
AB  =  BE  =  DE  (?). 


F1 


Statement :  AB  :  AF1  =  AF1 : 
Proof:  AB  is  tangent  to  the  O  (?). 
.'.  AE  :  AB  =  AB  :  AD  (?)  (335). 

Hence,  AE  +  AB  :  AE  =  AB  +  AD  :  AB  (?)  (294) . 

But  AE  +  AB  =  BFr,  and  ^4B  +  AD  =  AE  =  AF'  (Const.). 

/.  BF1  :  AF'  =  AF'  :  AB  (Ax.  6). 

That  is,  AB  :  AF'  =  AF1  :  BFr.  Q.E.D. 

365.   The  lengths  of  the  several  lines  of  363  and  364  may 
be  found  by  algebra,  if  the  length  of  AB  is  known. 
Thus,  if  AB  =  a,  we  know  in  363,  a  :  x  =  x  :  a  —  x. 
Hence,  a?  =  az  —  ax.      Solving  this  quadratic, 
a;  =  ^F=Ja(V5-l);  also,  a-  x  =  BF  =  \a  (3-  VS). 
Likewise,  in  364,  if  AB  =  a,  AF'  =y,  a:y  =  y:a+y. 
Solving  for  y,    y  =  AF'  =  \a  (V5  + 1). 
Also  a  +  y  =  BF1  =  \a  (3  +  V5). 


184  PLANE   GEOMETRY 


ORIGINAL   CONSTRUCTIONS 

1.  To  construct  a  fourth  proportional  to  lines  that  are  exactly  3  in., 

5  in.,  and  6  in.  long.     How  long  should  this  constructed  line  be  ? 

2.  To  construct  a  mean  proportional  between  lines  that  are  exactly 
4  in.  and  9  in.     How  long  should  this  constructed  line  be? 

3.  Will  a  fourth  proportional  to  three  lines  5  in.,  8  in.,  and  10  in.,  be 
the  same  length  as  a  fourth  proportional  to  5  in.,  10  in.,  and  8  in.  ?     to  8 
in.,  10  in.,  and  5  in.?  to  10  in.,  5  in.,  and  8  in.  ? 

4.  To  construct  a  third  proportional  to  lines  that  are  exactly  3  in.  and 

6  in.  long. 

5.  To  produce  a  given  line  A B  to  point  P,  such  that  AB  :  AP  =  3:5. 
[Divide  AB  into  three  equal  parts,  etc.] 

6.  To  divide  a  line  8  in.  long  into  two  parts  in  the  ratio  of  5:7. 
[Divide  the  given  line  into  12  equal  parts.] 

7.  To  solve  No.  6  by  constructing  a  triangle.     [See  308.] 

8.  To  divide  one  side  of  a  triangle  into  segments  proportional  to  the 
other  two  sides. 

9.  To  divide  one  side  of  a  triangle  externally  into  segments  propor- 
tional to  the  other  sides. 

10.  To  construct  two  straight  lines  having  given  their  sum  and  ratio. 
[Consult  No.  6.] 

11.  To  construct  two  straight  lines  having  given  their  difference  and 
ratio.     [Consult  No.  5.] 

12.  To  construct  a  triangle  similar  to  a  given  triangle  and  having  a 
given  perimeter.     [First,  use  358.] 

13.  To  construct  a  right  triangle  having  given  its  perimeter  and  an 
acute  angle.     [Constr.  a  rt.  A  having  the  given  acute  Z.    Etc.] 

14.  To   construct   a   triangle    having   given    its   perimeter   and   two 
angles.     [Constr.  a  A  having  the  two  given  /i.    Etc.] 

15.  To  construct  a  triangle  similar  to  a  given  triangle  and  having  a 
given  altitude. 

16.  To  construct  a  rectangle  similar  to  a  given  rectangle  and  having 
a  given  base. 

17.  To  construct  a  rectangle  similar  to  a  given  rectangle  and  having 
a  given  perimeter. 

18.  To  construct  a  parallelogram  similar  to   a  given  parallelogram 
and  having  a  given  base. 


BOOK    III 


185 


Through  P  draw  R  T  \\  to  OF. 


19.  To  construct  a  parallelogram  similar  to  a  given  parallelogram 
and  having  a  given  perimeter. 

20.  To  construct  a  parallelogram  similar  to  a  given   parallelogram 
and  having  a  given  altitude. 

21.  Three  lines  meet  at  a  point ;  it  is  required  to  draw  through  a 
given   point   another   line,    which    will    be 

terminated  by  the  outer  two  and  be  bisected 
by  the  inner  one. 

Construction :  From  E  on  BD  draw  Us. 
Etc. 

Statement :    RS  =  ST. 

22.  To  inscribe  in  a  given  circle  a  triangle  similar  to  a  given  triangle. 
Construction  :   Circumscribe  a  O  about  the  given  A;  draw  radii  to  the 

vertices;  at  center  of  given  O  construct  3  A  =  to  the  other  central  angles. 

23.  To  circumscribe  about  a  given  circle  a  triangle  similar  to  a  given 
triangle. 

Construction :   First,  inscribe  a  A  similar  to  the  given  A. 

24.  To  construct  a  right  triangle,  having  given  one  leg  and  its  projec- 
tion upon  the  hypotenuse. 

25.  To  inscribe  a  square  in  a  given  semicircle. 
Construction :    At  B  erect  BD  _L  to  .4  B  and  =  A  B 

Q at  R -,  dr-dwRU  \\toBD.   Etc.     Statement:   (?). 
Proof  :    RSTU  is  a  rectangle  (?). 
A  CRU  is  similar  to  A  CBD  (?). 
:.CU:CB  =  RU-.BD  (?).   But  CB  =  \ BD  (?) . 
/.  CU=\RU  (<>).   Etc. 

26.  To   inscribe  in   a  given  semicircle,  a  rec- 
tangle similar  to  a  given  rectangle. 

Construction  :   From  the  midpoint  of  the  base 
draw  line  to  one  of  the  opposite  vertices.     At  given  center  construct  an 
=  the  Z  at  the  midpoint.      Proceed  as  in  No.  25. 

Proof  :    First,  prove  a  pair  of  A  similar.    Etc. 

27.  To  inscribe  a  square  in  a  given  triangle.  ^ 
Construction :    Draw   altitude    A  D ;     con- 
struct the  square  A  DEF  upon  AD  as  a  side  ; 

draw  BF  meeting  AC  at  R. 

Draw  RU  \\  to  AD]   RS  \\  to  BC.  Etc. 

Statement:     (?).      Proof:     &BRU   and 
BFE  are  similar  (V).    Also  A  BRS  and  BAF  (?). 

Thence  show  that  SR  =  RU.    Etc. 


draw  DC,  meeting 
.0 


ou  c 


186  PLANE   GEOMETRY 

28.  To  inscribe  in  a  given  triangle  a  rectangle  similar  to   a  given 
rectangle. 

Construction :    Draw   the   altitude.     On  this  construct  a  rectangle 
similar  to  the  given  rectangle. 
Proceed  as  in  No.  27. 

29.  To  construct  a  circle  which  shall  pass 
through  two  given  points  and  touch   a  given 
line. 

Given  :   Points  A  and  B ;  line  CD. 

Construction:   Draw  line  AB  meeting   CD  ^~~ P  R 

at  P.  Construct  a  mean  proportional  between  PA  and  PB  (by 361). 
On  PD  take  PR  =  this  mean.  Erect  OR  JL  to  CD  at  R,  meeting  ±  bi- 
sector of  A  B  at  0.  Use  0  as  center,  etc. 

30.  To  construct  a  line  =  V2  in.    [Diag.  of  square  whose  side  is  1  in.] 

31.  To  construct  a  line  =  VS^in. 

[Hyp.  of  a  rt.  A,  whose  legs  are  1  in.  and  2  in.  respectively.] 

32.  To  divide  a  line  into  segments  in  the  ratio  of  1  :  V2. 

33.  To  divide  a  line  into  segments  in  the  ratio  of  1  :  \/5. 

34.  To  construct  a  line  as,    if  x  =  — ,  and  a,  b,  c  are  lengths  of  three 
given  lines.     [That  is,  to  construct  a;,  if  c  :  a  =  b  :  x  (291).] 

35.  To  construct  a  line  x,  if  x  =  — .     [3c :  a  =  b  :  x.'] 

oc 

36.  To  construct  a  line  a:,  if  a:  =  Vab.     [a  :  x  =  x :  &.] 

37.  To  construct  a  line  x,  if  x  =  — . 

c 

38.  To  construct  a  line  x,  if  x  =  Va2  _  52      [a  +  b:x  =  x:a —  b.~] 


39.  To  construct  a  line  x,  if  x  =  - — . 

c 

40.  To  construct  a  line  y,  if  ay  =  2  b2. 

41.  To  construct  a  line  =  VlO  in. 

42.  To  construct  a  line  =  2\/6  in. 

43.  To  construct  a  line  =  Va2  +  &2,  if  a  and  b  are  given  lines. 


BOOK    IV 


UNIT   OF    LENGTH 


AREAS 

366.  The  unit  of  surface  is  a  square  whose 
sides  are  each  a  unit  of  length. 

Familiar  units  of  surface  are  the  square  inch,  the 
square  foot,  the  square  meter,  etc. 

367.  The  area  of  a  surface  is  the  number  of 
units  of  surface  it  contains.     The  area  of  a  sur- 
face is  the  ratio  of  that  surface  to  the  unit  of 
surface. 

Equivalent  (=0=)  figures  are  figures  having  equal  areas. 

NOTE.  It  is  often  convenient  to  speak  of  "triangle,"  "rectangle," 
etc.,  when  one  really  means  "  the  area  of  a  triangle,"  or  "  the  area  of  a 
rectangle,"  etc. 

368.  THEOREM.   If  two  rectangles  have  equal  altitudes,  they  are  to 
each  other  as  their  bases. 

Given :  Rectangles  AC  and   D  c    H 

EG   having  =  altitudes,    and 
their  bases  being  AB  and  EF. 

To  Prove : 

AC  :  EG  =  AB  :  EF. 

n 

Proof:  I.   If  A  Band  EF  are  commensurable. 

There  exists  a  common  unit  of  measure  of  AB  and  EF 
(238).  Suppose  this  unit  is  contained  3  times  in  AB  and 
5  times  in  EF.  Hence,  AB  :  EF=  3  :  5  (Ax.  3). 

Draw  lines  through  these  points  of  division,  J_  to  the  bases. 
These  will  divide  rectangle  AC  into  three  parts  and  EG  into 
5  parts,  and  all  of  these  eight  parts  are  equal  (?)  (140). 

Hence,  AC  :  EG  =  3  :  5  (?). 

.'.  AC  :  EG  =  AB  :  EF  (Ax.  1).  Q.E.D. 

187 


188 


PLANE   (JKOMETRY 


II.    If  AB  and  EF  are  incom-     D  c    H  SG 

mensurable. 

There  does  not  exist  a  com- 
mon unit  (?)  (238).  Divide 
AB  into  several  equal  parts. 
Apply  one  of  these  as  a  unit 
of  measure  to  EF.  There  will  be  a  remainder,  RF  left 

over  (Hyp.)-     Draw  RS  _L  to  EF.     Now,  —  =  —   (Case  I). 

E8       ER 

Indefinitely  increase  the  number  of  equal  parts  of  AB ; 
that  is,  indefinitely  decrease  each  part,  or  the  unit  or 
divisor.  Hence  the  remainder,  RF,  will  be  indefinitely  de- 
creased (?). 

That  is,  RF  will  approach  zero  as  a  limit, 
and,  RFGS  will  approach  zero  as  a  limit. 
Hence,  ER  will  approach   EF  as  a  limit  (?), 
and   E8  will  approach  EG  as  a  limit  (?). 
A  C1  AC1 

Therefore,  -  -  will  approach  -  -  as  a  limit  (?), 

ES  EG 

and  —  will  approach  —  as  a  limit  (?). 

ER  EF 


.  £u_£±f  ,?.    (242). 

EG       EF 


Q.E.D. 


369.  THEOREM.    Two  rectangles  having  equal  bases  are  to  each 
other  as  their  altitudes.    (Explain.) 

370.  THEOREM.    Any  two  rectangles  are  to  each  other  as  the  prod- 
ucts of  their  bases  by  their  altitudes. 

F 


Given :  Rectangles  A  and  B  whose  altitudes  are  a  and  c 
and  bases  b  and  d  respectively. 


BOOK   IV 


189 


,ful 

i 


To  Prove  :  A  :  B  =  a  -  b  :  c  •  d. 

Proof :   Construct  a  third  rectangle  X  whose  base  is  b  and 
whose  altitude  is  c. 

Then  A  :  X=  a  :  c  (?)  (369).    Also,  X  :  B  =  b  :  d  (?). 
Multiplying,  A  :  B  =  a  -  b  :  c  -  d  (?)  (Ax.  3).  Q.E.D. 

371.  THEOREM.    The  area  of  a  rectangle  is  equal  to  the  product  of 
its  base  by  its  altitude. 

Given :  Rectangle  Zf,  whose 
base  is  b  and  altitude,  h. 

To  Prove :  Area  of  B  =  b  •  h. 

Proof :  Draw  a  square  J7,  each 
of  whose  sides  is  a  unit  of 
length.  This  square  is  a  unit  of  surface  (366). 

Now,  -  =  ^  =  b  -  h  (370).     But  -  =  area  of  B  (367). 

.-.  area  of  R  =  b  -  h  (Ax.  1).  Q.E.D. 

372.  THEOREM.   The  area  of  a  square  is  equal  to  the  square  of 
its  side.    (See  371.) 

373.  THEOREM.    The  area  of  a  parallelogram  is  equal  to  the  product 
of  its  base  by  its  altitude. 

Given  :  O  ABCD  whose  base  is  b  and  altitude,  h. 

To  Prove  :   Area  of  ABCD  =  b-  h. 

Proof :  From  A  and  B,  the  extremities  of  the  base,  draw 
Js  to  the  upper  base  meeting   F         D  EC 

it  in  F  and  E  respectively. 

In  rt.  A  ADF  and  BCE, 
AF  =  BE  (?),  AD  =  BC  (?). 
.'.  A  ADF=  A  BCE  (?). 


Now,  from  the  whole  figure  A 

take  A  ADF  and  parallelogram  ABCD  remains ;  and  from  the 
whole  figure  take  A  BCE  and  rectangle  ABEF  remains. 

.'.  n  ABCD  =0  rect.  ABEF  (Ax.  2). 

Rect.  ABEF  =  b  •  h  (?).    .-.  O  ABCD  =  b  -  h  (Ax.  1).  Q.E.D. 


190 


PLANE   GEOMETRY 


374.  COR.    All  parallelograms  having  equal  bases  and  equal  alti- 
tudes are  equivalent. 

375.  THEOREM.    Two  parallelograms  having  equal  altitudes  are  to 
each  other  as  their  bases. 

Proof:   P  =  b-h:   p'  =  b'-h  (?). 

•••?=fcH<A*-8>: 

376.  THEOREM.   Two   parallelograms  having  equal  bases  are  to 
each  other  as  their  altitudes. 

Proof:   CO- 

377.  THEOREM.   Any  two  parallelograms  are  to  each  other  as  the 
products  of  their  bases  by  their  altitudes. 

Proof:   (?). 

378.  THEOREM.   The  area  of  a  triangle  is  equal  to  half  the  product 
of  its  base  by  its  altitude. 

Given  :  A  ABC;  base  =  b ;    R 
altitude  =  h. 

To  Prove : 

Area  of  A  ABC  =  ^  b  •  h. 

Proof :    Through  A  draw 
AR  II  to  BC  and  through  C 

draw  CR  II  to  AB,  meeting  AR  at  R.     Now  ABCR  is  a  O  (?). 
Area  O  ABCR  =b-h  (?).      J  O  ABCR  =  l  b  -  h    (Ax.  3). 

.'.AABC=%b  •  h  (Ax.  1).  Q.E.D. 

379.  COR.    If  a  parallelogram  and  a  triangle  have  the  same  base 
and  altitude,  the  triangle  is  equivalent  to  half  the  parallelogram. 

380.  COR.    All  triangles  having  equal  bases  and  equal  altitudes  are 
equivalent. 

381.  COR.    All  triangles  having  the  same  base  and  whose  vertices 
are  in  a  line  parallel  to  the  base  are  equivalent  (?). 


BOOK    IV  191 

382.   THEOREM.   Two  triangles  having  equal  altitudes  are  to  each 
other  as  their  bases. 

Proof  :  A  T  =  J  b  -  h  ;  A  T1  =  %  b'h  (?). 

A  T  _  ^  b  h  _  b     ^x 
11/7       TT  v  )' 


'AT' 

383.  THEOREM.  Two  triangles  having  equal  bases  are  to  each  other 
as  their  altitudes. 

Proof:   (?). 

384.  THEOREM.   Any  two  triangles  are  to  each  other  as  the  products 
of  their  bases  by  their  altitudes. 

385.  THEOREM.   The  area  of  a  right  triangle  is  equal  to  half  the 
product  of  the  legs. 

386.  THEOREM.   The  area  of  a  trapezoid  is  equal  to  half  the  product 
of  the  altitude  by  the  sum  of  the  bases. 

Given:  Trapezoid   ABCD; 
altitude  =  h ;  bases  =  b  and*?. 

To  Prove :  / 

Area  ABCD  =  J  h  •  (b  +  <?). 

Proof :  Draw  diagonal  AC. 
Now  consider  the  A  ABC  and 

ADC  as  having  the  same  altitude,  A,  and  their  bases  b  and  <?, 
respectively. 

Now,       A  ABC 
and  A  ADC 


Adding,      A  ABC  +  AADC  =  ±b- h-\-±c>h     (Ax.  2). 

That  is,       trapezoid  ABCD  =  J  h  •  (6  +  c~)     (Ax.  6).      Q.E.D. 

387.   THEOREM.    The  area  of  a  trapezoid  is  equal  to  the  product  of 
the  altitude  by  the  median. 

Proof :   Area  ABCD  =  J  h  •  (b  +  c)  =  h  •  J  (5  +  <?)• 

But  J  (&  +  <?)  =  median  (144). 

Hence,  area  of  trapezoid  ABCD  =  h  •  m    (Ax.  6).       Q.E.D. 


192 


PLANE   GEOMETRY 


388.  THEOREM.  If  two  triangles  have  an  angle  of  one  equal  to  an 
angle  of  the  other,  they  are  to  each  other  as  the  products  of  the  sides 
including  the  equal  angles. 

Given:    A    ABC  and    DEF, 


To  Prove  : 

AABC      AB 


AC 


A  DEF      DE  -  DF 

Proof  :  Superpose  A  ABC 
upon  A  DEF  so  that  the  equal 
A  coincide  and  BC  takes.  the 
position  denoted  by  GH.  Draw  GF. 

Now  A  DGH  and  DGF  have  the  same  altitude  (a  _L  from  G 
to  DF),  and  A  DGF  and  DEF  have  the  same  altitude  (a  _L 
from  F  to  DE)  . 


A  DGF      DF 
Multiplying, 


A  DEF      DE 


DGH   DH 


That  is, 


A  DEF   DE  -  DF 
AABC  _AB  -  AC 
A  DEF   DE  •  DF 


(Ax.  6). 


Q.E.D. 


Ex.  1.  Prove  theorem  of  386  by  drawing  through  C  a  line  parallel  to 
AD,  dividing  the  trapezoid  into  a  parallelogram  and  a  triangle. 

Ex.2.  Which  includes  the  other,  the  word  "equal"  or  the  word 
"  equivalent "  ?  Which  of  these  words  conveys  no  idea  of  shape  ? 

Ex.  3.  What  is  the  area  of  a  parallelogram  whose  base  is  8  inches 
and  altitude  is  5  inches?  What  is  the  area  of  a  triangle  having  the  same 
base  and  altitude  ? 

Ex.  4.  Is  the  area  of  a  triangle  equal  to  half  the  base  multiplied  by 
the  whole  altitude?  Or  half  the  altitude  multiplied  by  the  whole  base? 
Or  half  the  base  multiplied  by  half  the  altitude?  Or  half  the  product 
of  the  base  by  the  altitude  ? 

Ex.  5.  If,  in  the  figure  of  388,  one  triangle  is  four  times  as  large  as 
the  other,  AB  =  10,  A  C  =  6,  DE  =  16,  find  DF. 

Ex.  6.  The  base  of  a  triangle  is  20  and  its  altitude  is  15.  The  bases 
of  an  equivalent  trapezoid  are  13  and  11;  find  its  altitude. 


HOOK    IV 


193 


389.   THEOREM.   Two    similar  triangles  are  to  each  other  as  the 
squares  of  any  two  homologous  sides. 


Given  :  Similar  A  ABC  and  DEF. 
To  Prove  :  —  =  ==  =  =z  = 

A  DEF      DW      DF* 

Proof:  ONE  METHOD.  Z  BAC  =  ^EDF  (?)  (323,1). 
AABC  =  AB.AC(?,^. 

A  7>#F       DE  -  DF 

A  ABC      AB      AC 


.    . 
That  is, 


XT         AC      AB   s<>\ 
=--x  —  .     Now,  —  =  --(?). 
A  DEF       DE       DF  DF       DE 


2 


A  ABC       AB^AB       AB      ,  A  v    ^ 
.*.  -          —  =  -  X  -  =         -    (  A.X.  u  ). 
A  DEF       DE      DE       j)E 

But  45  =  ^=*Zf?),   and  §  =  5  =  S 

DE      DF       EF  DE        DF        EF 


(297). 


A  ABC      AB 


A  DEF 


AC_  =  BCL    (Ax.  1). 


DF 


Q.E.D. 


ANOTHER  METHOD.    Denote  a  pair  of  homologous  alti 
tudes  by  h  and  h',  and  the  corresponding  bases  by  b  and  b1  '. 


A  DEF 
Hence, 


But       = 


That  is, 


A  DEF     b'      b'      b' 
A  ABC      liC2      AB2 


(322). 


Q.E.D. 


A  DEF      EF* 
BOBBINS'  PLANE  GEOM.  — 13 


DF 


194 


PLANE   GEOMETRY 


390.   THEOREM.   Two  similar  polygons  are  to  each  other  as  the 
squares  of  any  two  homologous  sides. 

B1 

> 

B 


E  D  E- 

Given  :  Similar  polygons  ABODE  and  A'B'C'D'E'. 
To  Prove :  ABODE :  A'B'C'D'E'  =  AB? :  A's'2  =  etc. 

Proof:  Draw  from  homologous  vertices,  A  and  A1,  all  the 
pairs  of  homologous  diagonals,  dividing  the  polygons  into  A. 
These  A  are  similar,  in  pairs  (?)  (327). 


AR 


A  R'      A'B' 


and 


CD 


AB 


AS' 


(?)  and 


A  T 

AT' 


AB 


AR  __    A.S  _  AT   ,A 

—     ~  7    I   /i  A.  • 


A 
+  A  S   +AT 


AB   (?)  (g()1)- 

AJZ'  v 


Therefore, 
Hence,  — - 

But 

.    polygon     ABODE 
'  Polyg°n  A'B'C'D'E'  ~  J7/?2  ~"  Wo*2 


AA_._^!_  m 

AR'       ^W2^'}' 


Q.E.D. 


Ex.  1.    The  base  of  a  triangle  is  6.     Find  the  base  of  a  similar  tri- 
angle that  is  9  times  as  large.     Five  times  as  large. 

Ex.  2.    The  area  of  a  polygon  is  104  and  its  longest  side  is  12.     What 
is  the  area  of  a  similar  polygon  whose  longest  side  is  15? 


BOOK    IV 


195 


H 


391.  THEOREM.   The  square  described  upon   the  hypotenuse  of  a 
right  triangle  is  equivalent  to  the  sum  of  the  squares  described  upon 
the  legs. 

Given:    (?). 

To  Prove:  (?). 

Proof  :    Draw    CL  _L  to 
AB,  meeting  AB  at  K  and    F< 
ED  at  L.   Draw  BF  and  CE. 

Now,    AACB,  ACG,  and 
BCH&i-e  all  rt.  A  (?). 

Hence,   ACH    and    BCG 
are  straight  lines  (?)  (45). 

Also,  AELK  and  BDLK 
are  rectangles  (?)  (127). 

In  A    ABF     and     ACE, 
AB=AE,    AF  —  AC  (128), 

and  /_BAF=  /_  CAE.  (Each 
=  a  rt.Z  +  Z  BAG.) 

\.  *  s      \.  J  * 

Also,  A4.B.Fand  square  AG  have  the  same  base,  AF,  and 
the  same  altitude,  AC. 

Hence,    square   AG  o  2  A  ABF  (379). 

Similarly,  rectangle  AKLE  =c=  2  A  ACE  (?). 

Therefore,  rectangle  AKLE  =^  square  ACGF  (Ax.  1). 

By  drawing  ^/  and  CD,  it  is  proven  in  like  manner,  that 
rectangle  BDLK  =c=  square  BCHI.     Then,  by  adding, 
square  ABDE  =c=  square  ACGF+  square  BCHI  (Ax.  2).    Q.E.D. 

392.  THEOREM.  The  square  described  upon  one  of  the  legs  of  a  right 
triangle  is  equivalent  to  the  square  described  upon  the  hypotenuse 
minus  the  square  described  upon  the  other  leg.    (Explain.) 


Ex.  1.  If  the  legs  of  a  right  triangle  are  15  and  20,  what  is  the  hypote- 
nuse ?  If  the  legs  are  »i2  -  1  and  2  m,  what  is  the  hypotenuse  ? 

Ex.  2.  What  is  the  difference  in  the  wording  of  the  theorems  of  343 
and  391  ?  Which  proof  is  purely  algebraic  ?  Which  is  geometric  ? 


196 


PLANE   GEOMETRY 


393.  THEOREM.  If  the  three  sides 
of  a  right  triangle  are  the  homolo- 
gous sides  of  three  similar  polygons, 
the  polygon  described  upon  the  hypote- 
nuse is  equivalent  to  the  sum  of  the 
two  polygons  described  upon  the  legs. 

Proof:    -  ==  (?). 
^        AB2 


And 


-==-(?).    Adding, 


rt       |         >TT  A  /"i*      -I-       f> C<  ~~A~I? 

^  =  AC—  9  B°  =^=1.     (Explain.) 


.'.R  o  8+  r(?). 


Q.E.D. 


394.  COR.    If  the  three  sides  of  a  right  triangle  are  the  homolo- 
gous sides  of  three  similar  polygons,  the  polygon  described  upon  one 
of  the  legs  is  equivalent  to  the  polygon  described  upon  the  hypote- 
nuse minus  the  polygon  described  upon  the  other  leg. 

395.  THEOREM.      The    two 
squares  described  upon  the  legs 
of  a  right  triangle  are  to  each 
other  as  the  projections  of  the 
legs  upon  the  hypotenuse. 

Proof:  square  s  = 

Square   T 

AB  •  AP       AP        ,„      ,    . 

—  = (Explain. 

AB • BP       BP 


396.  THEOREM.  If  two  similar  poly- 
gons are  described  upon  the  legs  of  a 
right  triangle  as  homologous  sides,  they 
are  to  each  other  as  the  projections  of 
the  legs  upon  the  hypotenuse. 

^2 


Proof:   -=^-  = 


AP 
BP 


(Explain.) 


P  B 


BOOK    IV  197 


ORIGINAL   EXERCISES   (THEOREMS) 

1.    If  one  parallelogram  has  half  the  base  and  the  same  altitude  as 
another,  the  area  of  the  first  is  half  the  area  of  the  second. 

2    If  one  parallelogram    has  half  the  base  and  half  the  altitude  of 
another,  its  area  is  one  fourth  the  area  of  the  second. 

3.  State  and  prove  two  analogous  theorems  about  triangles. 

4.  If  a  triangle  has  half  the  base  and  half  the  altitude  of  a  paral- 
lelogram, the  triangle  is  one  eighth  of  the  parallelogram. 

5.  The  area  of  a  rhombus  is  equal  to  half  the  product  of  its  diagonals. 

6.  The  diagonals  of  a  parallelogram  divide  it  into  four  equivalent 
triangles. 

7.  The  diagonals  of  a  trapezoid  divide  it  into  four  triangles,  two  of 
which  are  similar  and  the  other  two  are  equivalent. 

8.  If  a  parallelogram  has  half  the  base  and  half  the  altitude  of  a 
triangle,  its  area  is  half  the  area  of  the  triangle. 

9.  The  line  joining  the  midpoints  of  two  sides  of  a  triangle  forms  a 
triangle  whose  area  is  one  fourth  the  area  of  the  original  triangle. 

10.  The  line  joining  the  midpoints  of  two  adjacent  sides  of  a  paral- 
lelogram cuts  off  a  triangle  whose  area  is  one  eighth  of  the  area  of  the 
parallelogram. 

11.  If  one  diagonal  of  a  quadrilateral  bisects     AT B 

the  other,  it  also  divides  the  quadrilateral  into 

two  equivalent  triangles. 
To  Prove :  A  ABC  -  A  ADC.  D 

12.  Either  diagonal  of  a  trapezoid  divides  the  figure  into  two  triangles 
whose  ratio  is  equal  to  the  ratio  of  the  bases  of  the  trapezoid.     Prove 
two  ways.     [By  382  and  by  388.] 

13.  If,  in  triangle  ABC,  D  and  E  are  the  midpoints  of  sides  AB  and 
A C  respectively,  &BCD-&BEC. 

14.  If  the  diagonals  of  quadrilateral  AB  CD  meet  at  E  and  A  ABE  ~ 
A  CDE,  the  sides  A  D  and  BC  are  parallel.     [Prove  A  ABD  ^  A  A  CD.'] 

15.  The  square  described  upon  the  hypotenuse  of  an  isosceles  right 
triangle  is  equivalent  to  four  times  the  triangle. 

16.  The  square  described  upon  the  diagonal  of  a  square  is  double  the 
original  square. 


198 


PLANK   GEOMETRY 


17.  Any  two  sides  of  a  triangle  are  reciprocally  proportional  to  the 
altitudes  upon  them.     [Use  378  and  291.] 

18.  In  equivalent  triangles  the  bases  and  the  altitudes  upon  them  are 
reciprocally  proportional. 

19.  If  two  isosceles  triangles  have  the  legs  of 
one  equal  to  the  legs  of  the  other,  and  the  vertex- 
angle    of  the  one  the  supplement  of  the  vertex- 
angle  of  the  other,  the  triangles  are  equivalent. 

Given  :  &  ABC  and  A  CD,  etc. 


20.  Two  triangles  are  equivalent  if  they 
have  two  sides  of  one  equal  to  two  sides  of 
the  other  and  the  included  angles  supplemen- 
tary. 

Proof:  Z.CAD  =  Z  C'AD'  (?)  and  CA  = 
C'A  (V).  .%  the  rt.  &  are  =  (?).  Etc. 


C' 


21.  If  two  triangles  have  an  angle  of  one  the 
supplement  of  an  angle  of  the  other,  the  triangles 
are  to  each  other  as  the  products  of  the  sides  in- 
cluding these  angles.  C" 

Given :  &  ABD  and  EEC,  A  at  B  supplementary. 

Proof :   Draw  DC,  use  A  BCD,  and  proceed  as  in  388. 


22.  The  area  of  a  triangle  is  equal  to  half 
the  perimeter  of  the  triangle  multiplied  by  the 
radius  of  the  inscribed  circle. 

Proof:  Draw  CM,  etc.  &AOC  =  $AC  •  r(?); 
&AOB  =  \AB  -  r  (?),  etc.  Add. 


23.  The  area  of  a  polygon  circumscribed  about  a  circle  is  equal  to 
half  the  product  of  the  perimeter  of  the  polygon  by  the  radius  of  the 
circle. 

24.  The  line  joining  the  midpoints  of  the  bases  of  a  trapezoid  bisects 
the  area  of  the  trapezoid. 

25.  Any  line  drawn  through  the  midpoint  of  a  diagonal  of  a  paral- 
lelogram, intersecting  two  sides,  bisects  the  area  of  the  parallelogram. 

26.  The  lines  joining  (in  order)  the  midpoints  of  the  sides  of  any 
quadrilateral  form  a  parallelogram  whose  area  is  half  the  area  of  the 
quadrilateral. 


BOOK   IV  199 

27.  If  any  point  within  a  parallelogram  is  joined  to  the  four  vertices, 
the  sura  of  one  pair  of  opposite  triangles  is  equivalent  to  the  sum  of  the 
other  pair;  that  is,  to  half  the  parallelogram. 

28.  Is  a  triangle  bisected  by    an  altitude?     By  the  bisector  of  an 
angle?     By  a  median?     By  the  perpendicular  bisector  of  a  side?     Give 
reasons. 

29.  If  the  three  medians  of  a  triangle  are 
drawn,  there  are  six  pairs  of  triangles  formed, 
one  of  each  pair  being  double  the  other. 

To  Prove:  A  A  OB  =  2  A  A  OE;  etc. 

M V* 

30.  If  the  midpoints  of  two  sides  of  a  tri- 
angle are  joined  to  any  point  in  the  base,  the  quadrilateral  formed  is 
equivalent  to  half  the  original  triangle. 

31.  If   lines  are  drawn  from  the  midpoint 
of  one  leg  of  a  trapezoid  to  the  ends  of  the 
other  leg,  the  middle  triangle  thus  formed  is 
equivalent  to  half  the  trapezoid. 

Proof:  Draw  median  EF  =  m.  Then  EF 
is  li  to  the  bases  (?).  Denote  the  altitude  of 
the  trapezoid  by  h.  Then  EF  bisects  h  (?).  A  BFE  =  |  m  -  \  h  (?). 

A  AEF  =  \  m  .  \h  (?).     /.A  ABE  =  \  mh.     Consult  387. 

32.  The  area  of   a  trapezoid  is  equal  to  the  product  of  one  of  the 
non-parallel  sides,  by  the  perpendicular  upon  it  from  the  midpoint  of  the 
other. 

Proof :  Prove  that  A  ABE  =  half  the  trapezoid,  by  No.  31.     But  the 
A  A  BE=±ABx  the  _L  to  AB  from  E  (?). 

.-.  half  the  trapezoid  =  \  AB  x  this  J_  (Ax.  1).     Etc. 

33.  If  through  the  midpoint  of  one  of  the 
non -parallel    sides  of   a  trapezoid   a   line    is 
drawn  parallel  to  the  other  side,  the  parallelo- 
gram formed  is  equivalent  to  the  trapezoid. 

34.  If  two   equivalent  triangles  have  an 
angle  of  one  equal  to  an  angle  of  the  other, 

the  sides  including  these  angles  are  reciprocally  proportional. 

35.  The  sum  of  the  three  perpendiculars  drawn   to  the  three  sides 
of  an  equilateral  triangle  from  any  point  within  is  constant  (being  equal 
to  the  altitude  of  the  triangle). 

Proof:  Join  the  point  to  the  vertices.      Set  the  sum  of  the  areas  of 
the  three  inner  &  equal  to  the  area  of  the  whole  A.    Etc. 


£ E 

X 


200 


PLANE   GEOMETRY 


36.   In  the  figure  of  391,  prove  : 

(i)    Points  /,  (7,  and  F  are  in  a  straight  line. 

(it)   CE  and  BF  are  perpendicular. 
(in)   AG  and  EH  are  parallel. 


1 

3 

rn 

n 

m             m 

m 

m 

On 

n              n 

n 

m 

n 

[See  Ex.  20.] 

37.  The  sum  of  the  squares  described  upon 
the  four  segments  of  two  perpendicular  chords  in 
a  circle  is  equivalent  to  the  square  described  upon 
the  diameter.    (Fig.  is  on  page  178.) 

38.  The  square  described  upon  the  sum  of  two 
lines  is  equivalent  to  the  sum  of  the   squares  de- 
scribed upon  the  two  lines,  plus  twice  the  rectangle 
of  these  lines. 

To  Prove  :  Square  A  E  =0=  m2  +  n2  +  2  mn. 

39.  The  square  described  upon  the  difference  of 
two  lines  is  equivalent  to  the  sum  of  the  squares  de- 
scribed upon  the  two  lines  minus  twice  the  rectangle 
of  these  lines. 

To  Prove  :  Square  AD  =  m2  -f  n2  —  2  mn. 

40.  A  and  B  are  the  extremities  of  a  diameter 
of  a  circle  ;  C  and  D  are  the  points  of  intersection  of 
any  third  tangent  to  this  circle,  with  the  tangents 
at  A    and  B  respectively.     Prove  that  the  area  of 
ABDC  is  equal  to^AB-  CD. 

'41.  If  the  four  points  midway  between  the  center  and  vertices  of  a 
parallelogram  be  joined  in  order,  there  will  be  a  parallelogram  formed  ; 
it  will  be  similar  to  the  original  parallelogram  ;  its  perimeter  is  half  of 
the  perimeter  of  the  original  figure  ;  and  its  area  is  one  quarter  of  the 
area  of  the  original  figure. 

42.  If  two  equivalent  triangles  have  the  same  base  and  lie  on  opposite 
sides  of  it,  the  line  joining  their  vertices  is  bisected  by  the  base. 

43.  What  part  of  a  right  triangle  is  the  quadrilateral  which  is  cut 
from  the  triangle  by  a  line  joining  the  midpoints  of  the  legs? 

44.  Show  by  drawing  a  figure  that  the  square  on  half  a  line  is  one 
fourth  the  square  on  the  whole  line. 

45.  From  M,  a  vertex  of  parallelogram  LMNO,  a  line  MPX  is  drawn 
meeting  NO  at  P  and  LO  produced,  at  X.     LP  and  NX  are  also  drawn. 
Prove  triangles  LOP  and  XNP  are  equivalent. 


G 

-in 

n      n 

C 

n 

m-n 

m-n          m-n 

m-n 

m-n 

n 

n 

n 

m 

BOOK  IV 


201 


FORMULAS 

397.   PROBLEM.    To  derive  a  formula  for  the  area  of  a  triangle  in 
terms  of  its  sides. 

Given :  A  ABC,  having  sides 
=  a,  b,  c. 

Required:  To  derive  a  for- 
mula for  its  area,  containing 
only  a,  b,  and  c.  C 

Solution  :  *  Draw  altitude  AD. 

Now  CD  =  bpa=  a  +      ~'6    (349), 
2a 

and        AD2  =  Iff  -  CD2  (392). 


Hence,  h?  =  \b  +  —  ~\r~  '  \  (^v  fact°rrng)^ 


and 


_  Q  +  6  +  g)(a  +  b  —  c)  (c  +  a  —  b  )  (  c  —  a  + 
~-~ 


4a2 


Now,  area  of  A  =  -  a  -  h  = 
2 


?     .  /r«  +  J  +  c)(«  +  6  -  e)(«  -  6  +  c)  (  -  a  +  6  +  e) 
2     V-       -  4^- 

/.area  =  %  V(« 


.  Q.E.F. 


To  simplify  this  formula,  let  us  call  a  +  b  +  c=2s. 
Then,  it  is  evident  that  a  +  b  —  c=  2  (*  —  c»)  ; 
a  —  b  +  c=  2»  —  6)  ;    —  a  +  &  +  c  =  2(*  —  «). 


202  PLANE   GEOMETRY 

Substituting,  in  the  formula  for  area, 


Area  of  A  =  J  V2  «  .  2(«  -  c)  -  2(«  -  5)  .  2(s  -  a). 
That  is, 

Area  of  A  =  Vsfs  -  a)  (s  -  6)  (s  -c) . 


EXERCISE.   Find  the  area  of  a  triangle  whose  sides  are   17,  25,  28. 
Here,  a  =  17,  b  =  25,  c  =  28,  s  =  35,  s  -  a  =  18,  s  -  b  =  10,  s  -  c  =  7. 
Area  =  V35  •  18  . 10  .  7  =  Vf*~5*~^W2  =  210. 


398.   PROBLEM.    To  derive  formulas  for  the  altitudes  of  a  triangle 
in  terms  of  the  three  sides. 


Solution:  Area  =  |  aha  =  V*  (*  —  a)  (*  —  5)  (*  —  0). 


_  VsQ- 


Similarly,  ft,  = 


.  ft 


^ 


399.   PROBLEM.   To   derive  the  formulas  for  the  altitude  and  the 
area  of  an  equilateral  triangle,  in  terms  of  its  side. 

Solution:    Let  each  side  =  a, 
and  altitude  =  h. 

Then,  A2  =  a2  -  ^  =  —(?). 


Also, 
Area  =i  base  •  h  =  \a  -  ?  V3. 


2       2 


I 


Ex.  1.  Find  the  area  of  the  triangle  whose  sides  are  7,  10,  11. 

Ex.  2.  Find  the  area  of  the  triangle  whose  sides  are  8,  15,  17. 

Ex.  3.  Find  the  area  of  the  equilateral  triangle  whose  side  is  8. 

Ex.  4.  Find  the  side  of  the  equilateral  triangle  whose  area  is  121  V^ 

Ex.  5.  Find  the  area  of  the  equilateral  triangle  whose  altitude  is  10. 


BOOK   IV 


203 


400.   PROBLEM.    To  derive  the  formula  for  the  radius  of  the  circle 
inscribed  in  a  triangle,  in  terms  of  the  sides  of  the  triangle. 


Solution : 


Area  of  A  A  OB  =  | -c  -  r\ 
area  of  A  AOC  =  %b>r  (?). 
area  of  A  BOC  =    a- 


Adding,  area  of  AABC=^  (a  +  b  -f  c)  r  =  sr. 

(Because,  -|  (a  +  b  +  c)  =  s.) 

Hence,  r=areaofA^BC. 


.'.  r  = 


Vs  (s  -  a)  (8  -  ft)  (a  -  c) 


401.  PROBLEM.  To  derive  the 
formula  for  the  radius  of  the  circle 
circumscribed  about  a  triangle,  in 
terms  of  the  sides  of  the  triangle. 

Solution  : 

2fl  •  ha  =  b  .  c  (?)  (337). 

b  •  c 


R  = 


2A 


..JB   = 


a 


4^8(8-0)  (*- 


Ex.  1.  Find  the  radius  of  the  circle  inscribed  in,  and  the  radius  of  the 
circle  circumscribed  about,  the  triangle  whose  sides  are  17,  25,  28. 

Ex.  2.  Find  for  triangle  whose  sides  are  11,  14,  17,  the  radii  of  the 
inscribed  and  circumscribed  circles. 


'204:  PLANK   GEOMETRY 


ORIGINAL   EXERCISES  (NUMERICAL) 

1.  The  base  of  a  parallelogram  is  2  ft.  6  in.  and  its  altitude  is  1  ft. 
4  in.     Find  the  area.     Find  the  side  of  an  equivalent  square. 

2.  The  area  of  a  rectangle  is  540  sq.  in.  and  its  altitude  is  15  m. 
Find  its  base  and  diagonal. 

3.  The  base  of  a  rectangle  is  3  ft.  4  in.  and  its  diagonal  is  3  ft.  5  in. 
Find  its  area. 

4.  The  bases  of  a  trapezoid  are  2  ft.  1  in.,  and  3  ft.  4  in.,  and  the 
altitude  is  1  ft.  2  in.     Find  the  area. 

5.  The  area  of  a  trapezoid  is  736  sq.  in.  and  its  bases  are  3  ft.  and 
4  ft.  8  in.     Find  the  altitude. 

6.  The  area  of  a  certain  triangle  whose  base  is  40  rd.,  is  3.2  A.    Find 
the  area  of  a  similar  triangle  whose  base  is  10  rd.     Find  the  altitudes  of 
these  triangles. 

7.  The  base  of  a  certain  triangle  is  20  cm.     Find  the  base  of  a  simi- 
lar triangle  four  times  as  large ;   of  one  five  times  as  large ;   twice  as 
large;  half  as  large;  one  ninth  as  large. 

8.  The  altitude  of  a  certain  triangle  is  12  and  its  area  is  100.     Find 
the  altitude  of  a  similar  triangle  three  times  as  large.     Find  the  base  of 
a  similar  triangle  seven  times  as  large.     Find  the  altitude  and  base  of  a 
similar  triangle  one  third  as  large. 

9.  The  area  of  a  polygon  is  216  sq.  m.  and  its  shortest  side  is  8  m. 
Find  the  area  of  a  similar  polygon  whose  shortest  side  is  10  m.     Find  the 
shortest  side  of  a  similar  polygon  four  times  as  large;  one  tenth  as  large. 

10.  If  the  longest  side  of  a  polygon  whose  area  is  567  is  14,  what  is 
the  area   of   a  similar  polygon  whose   longest  side  is   12  ?   of  another 
whose  longest  side  is  21  ? 

11.  Find  the  area  of  an  equilateral  triangle  whose  sides  are  each  6  in. 
Of  another  whose  sides  are  each  10 V  3  ft. 

12.  Find  the  area  of  an  equilateral  triangle  whose  altitude  is  4  in. ; 
of  another  whose  altitude  is  18  dm. 

13.  The  area  of  an  equilateral  triangle  is  64 V3.    Find  its  side  and  its 
altitude. 

14.  The  area  of  an  equilateral  triangle  is  90  sq.  m.     Find  its  altitude. 

15.  Find  the  side  of  an  equilateral  triangle  whose  area  is  equal  to  a 
square  whose  side  is  15  ft. 


BOOK  IV  205 

16.  The  equal  sides  of  an  isosceles  triangle  are  each  17  in.  and  the 
base  is  16  in.     Find  the  area. 

17.  Find  the  area  of  an  isosceles  right  triangle  whose  hypotenuse  is 
2  ft.  6  in. 

18.  Find  the  area  of  a  square  whose  diagonal  is  20  m. 

19.  There  are  two  equilateral  triangles  whose  sides   are  33  and  56 
respectively.     Find  the  side  of  the  third,  equivalent  to  their  sura.     Find 
the  side  of  the  equilateral  triangle  equivalent  to  their  difference. 

20.  There  are  two  similar  polygons  two  of  whose  homologous  sides 
are  24  and  70.     Find  the  side  of  a  third  similar  polygon  equivalent  to 
their  sum  ;  the  side  of  a  similar  polygon  equivalent  to  their  difference. 

21.  What  is  the  area  of  the  right  triangle  whose  hypotenuse  is  29 
cm.  and  whose  short  leg  is  20  cm.  ? 

22.  The  base  of  a  triangle  is  three  times  the  base  of  an  equivalent 
triangle.     What  is  the  ratio  of  their  altitudes? 

23.  The  bases  of  a  trapezoid  are  56  ft.  and  44  ft.  and  the  non- paral- 
lel sides  are  each  10  ft.     Find  its  area.     Also  find  the  diagonal  of  an 
equivalent  square. 

24.  The  base  of  a  triangle  is  80  m.,  and  its  altitude  is  8  m.    Find  the 
area  of  the  triangle  cut  off  by  a  line  parallel  to  the  base  and  at  a  dis- 
tance of  3  m.  from  it.     Another,  cut  off  by  a  line  parallel  to  the  base 
and  6  m.  from  it. 

25.  The  bases  of  a  trapezoid  are  30  and  55, 
and  its  altitude  is  10.     If  the  non-parallel  sides 
are  produced  till  they  meet,  find  the  area  of  the 
less  triangle  formed. 

[The  &  are  similar.    .'.30:  55=ziz  +  10.  Etc.] 

26.  The  diagonals  of  a  rhombus  are  2  ft.  and  70  in.     Find  the  area; 
the  perimeter  ;  the  altitude. 

27.  The  altitude  (h)  of  a  triangle  is  increased  by  n  and  the  base  (&) 
is  diminished  by  x  so  the  area  remains  unchanged.     Find  x. 

28.  The  projections  of  the  legs  of  a  right  triangle  upon  the  hypote- 
nuse are  8  and  18.     Find  the  area  of  the  triangle. 

29.  In  triangle  ABC,  AB  is  5,  BC  is  8,  and  AB  is  produced  to  P, 
making  BP=Q.     BC  is  produced  (through  B)  to  Q  and  PQ  drawn  so  the 
triangle  BPQ  is  equivalent  to  triangle  ABC.      Find  the  length  of  BQ. 

[Use  388.] 


206  PLANE   GEOMETRY 

30.  The  angle  C  of  triangle  ABC  is  right;  AC  =  5;  EC  =  12.     BA 
is  produced  through  A,  to  Z>  making  AD  =  4;  CM  is  produced  through 
.4,  to  E  so  triangle  ^4£Z)  is  equivalent  to  triangle  ABC.     Find  A  E. 

31.  Find  the  area  of  a  square  inscribed  in  a  circle  whose  radius  is  6. 

32.  Find  the  side  of  an  equilateral  triangle  whose  area  is  25  V?. 

33.  Two  sides  of  a  triangle  are  12  and  18.      What  is  the  ratio  of  the 
two  triangles  formed  by  the  bisector  of  the  angle  between  these  sides? 

34.  The  perimeter  of  a  rectangle  is  28  m.  and  one  side  is  5  m.     Find 
the  area. 

35.  The  perimeter  of  a  polygon  is  5  ft.  and  the  radius  of  the  inscribed 
circle  is  5  in.     Find  the  area  of  the  polygon. 

In  the  following  triangles,  find  the  area,  the  three  altitudes,  radius  of 
inscribed  circle,  radius  of  circumscribed  circle : 

36.  a  =  13,  b  =  14,  c  =  15. 

37.  a  =  15,  b  =  41,  c  =  52. 

38.   20,  37,  51.  39.    25,  63,  74.  40.    140,  143,  157. 

41.  The  sides  of  a  triangle  are  15,  41,  52 ;  find  the  areas  of  the  two 
triangles  into  which  this  triangle  is  divided  by  the  bisector  of  the  largest 
angle. 

42.  Find  the  area  of  the  quadrilateral  ABCD  if  AB  =  78  m.,  BC  = 
104m.,  CD  =  50  m.,  AD  =  120  m.,  and  AC  =  130  m. 

43.  One  diagonal  of  a  rhombus  is  j8^  of  the  other  and  the  difference 
of  the  diagonals  is  14.     Find  the  area  and  perimeter  of  the  rhombus. 

44.  A  trapezoid  is  composed  of  a  rhombus  and  an  equilateral  triangle ; 
each  side  of  each  figure  is  16  inches.     Find  the  area  of  the  trapezoid. 

45.  Find  the  side  of  an  equilateral  triangle  equivalent  to  the  square 
whose  diagonal  is  15  \/2. 

46.  Which  of  the'figures  in  No.  45  has  the  less  perimeter? 

47.  In  a  triangle  whose  base  is  20  and  whose  altitude  is  12,  a  line  is 
drawn  parallel  to  the  base,  bisecting  the  area  of  the  triangle.    Find  the 
distance  from  the  base  to  this  parallel. 

48.  Parallel  to  the  base  of  a  triangle  whose  base  is  30  and  altitude  is 
18  are  drawn  two  lines  dividing  the  area  of  the  triangle  into  three  equal 
parts.     Find  their  distances  from  the  vertex. 

49.  Around  a  rectangular  lawn  30  yards  x  20  yards  is  a  drive  16  feet 
wide.    How  many  square  yards  are  there  in  the  drive  ? 


BOOK  TV 
CONSTRUCTIONS 


207 


402.   PROBLEM.    To  construct  a  square  equivalent  to  the  sum  of  two 
squares. 

Y 


Given  :  (?).  Required  :  (?).  Construction  :  Construct  a 
rt.  Z  E,  whose  sides  are  EX  and  EY.  On  EX  take  EF  =  AB, 
and  on  EY  take  EG— CD.  Draw  FG.  On  FG  construct  square  T. 

Statement :    T=C=  R  +  <s.  Q.E.F. 

Proof  :    ~GF2  =  EF2  +  ISG2  (?).        But  GF2  =  T  ;    ^F2  =  R  ; 

EG2  =  s  (?)  (372).     Hence,  r  =c=  u  +s  (Ax.  6).          Q.E.D. 

403.  PROBLEM.  To  construct  a  square  equivalent  to  the  sum  of 
several  squares. 

Given :  Squares  whose  sides  are 
a,  6,  <?,  d. 

Required:  To  construct  a  square 

=0=  a2  +  52  +  c2  4-  d2. 

Construction  :  Construct  a  rt.  Z 
whose  sides  are  equal  to  a  and  b. 
Draw  hypotenuse  BC.  At  B  erect 

a  _L  =  c  and  draw  hypotenuse,  DO.    Q 

At  D  erect  a  J_  =  d,  etc.  4 

Statement:    The   square    con-    c 

structed  on  EC  is  =0=  to  the  sum  of 
the  several  given  squares. 

Proof:    EC2  =DC2  +  d2  =  (^C2  +  e*2)  + 


Q.E.F. 


.  Q.E.D. 


208 


PLANE   GEOMETRY 


404.   PROBLEM.    To  construct  a  square  equivalent  to  the  difference 
of  two  given  squares. 


R 

S 

a   c           c 

Given:  (?).    Required:  (?). 

Construction :  At  one  end  of  indefinite  line,  EX,  erect  EG 
_L  to  EX  and  =  CD  (a  side  of  the  less  square,  s).  Using  G  as 
center  and  AB  as  radius,  describe  arc  intersecting  EX  at  F. 

Draw  GF.     On  EF  construct  square  T. 

Statement:  T^R—S.     Q.E.F.  Proof:  (?). 

405.  PROBLEM.   To   construct  a   polygon   similar  to  two  given 
similar  polygons  and  equivalent  to  their  sum. 

Construction:  Like  402.     Proof:   (393). 

406.  PROBLEM.   To  construct   a  polygon   similar  to  two   given 
similar  polygons  and  equivalent  to  their  difference. 

Construction:  Like  404.     Proof:   (394). 

407.  PROBLEM.   To  construct  a  square  equivalent  to  a  given  paral- 
lelogram. 


A  B      A1 

Given:  (?).     Required:  (?). 


M 


B'  E'    X  B' 

Construction :   Construct  a 


BOOK   IV 


209 


mean  proportional  between  the  base,  AB,  and  the  altitude, 
BE-,  on  this  mean  proportional,  B'G,  construct  a  square,  3f. 

Statement:    Square  M  ^  parallelogram  P.  Q.E.F. 

Proof  :    AB:BrG  =  B1  G  :  BE  (Const.). 

.'.  Wo2  ==  AB  •  BE  (?).      But  WG2  =  M  (?). 

And4B-B£ssP(?)(371),     Hence, M^P (Ax.  6).  Q.E.D. 

408.  PROBLEM.    To  construct  a  square  equivalent  to  a  given  tri- 
angle. 

Construct  a  mean  proportional  between  half  the  base  and 
the  altitude,  and  proceed  as  in  407. 

409.  PROBLEM.   To  construct  a  triangle  equivalent  to  a  given  poly- 
gon. 


A  B        G         i        A  G 

Given :  Polygon  AD.     Required  :    To  construct  a  A  =0=  AD. 

Construction  :  Draw  a  diagonal,  BD,  connecting  any  vertex 
(.B)  to  the  next  but  one  (-D).  From  the  vertex  between 
these  (C),  draw  CG  II  to  BD,  meeting  AB  prolonged,  at  G. 
Draw  DG.  Repeat  (2d  figure)  by  drawing  EG,  then  DH  II  to 
EG,  and  EH.  Repeat  again  by  drawing  AE,  FI II  to  AE,  and  El. 

Statement :  A  IEH  =0=  polygon  ABCDEF.  Q.E.F. 

Proof :  In  1st  fig.,  A  BGD  o  A  BCD  (381 ;  BD  is  the  base). 

Add          polygon  ABDEF  =  polygon  ABDEF. 

.'.  polygon  AGDEF  =0=  polygon  ABCDEF  (Ax.  2). 

Likewise,  A HEFoAGDEF',  and  AlEH=oAHEF.   (Explain.) 

.'.   A  IEHO  polygon  ABCDEF  ( Ax.  1).  Q.E.D. 

410.  PROBLEM.  To  construct  a  square  equivalent  to  a  given  poly- 
gon. [Use  409  and  408.] 

BOBBINS'  PLANE  GEOM.  — 14 


210 


PLANE   GEOMETRY 


411.   PROBLEM.   To    construct  a  parallelogram    (or  a  rectangle) 
equivalent  to  a  given  square,  and  having : 

I.  The  sum  of  its  base  and  altitude  equal  to  a  given  line. 
II.  The  difference  of  its  base  and  altitude  equal  to  a  given  line. 

x. H      C1  i 

E 


p.. 


G 


D' 


A  B     C          G  D 

1.  Given  :  Square  5  and  line  CD. 

Required  :  To  construct  a  O  =0=  S,  whose  base  +  altitude 
shall  =  CD. 

Construction  :  On  CD  as  a  diameter  describe  a  semicircle. 
At  c  erect  CE  _L  to  CD  and  =  AB.  Through  E  draw  EF  \\  to 
CD,  meeting  the  circumference  at  F.  Draw  FG  J_  to  CD. 
Take  G'D'  =  GD  and  draw  XT  II  to  G'D'  at  the  distance  from 
it  =  CG.  On  XY  take  HI  =  GD.  Draw  EG1  and  ID1. 

Statement  :  O  G'D'IH  =0=  s  and  base  +  alt.  =  CD.        Q.E.F. 


Proof:   G'D'lffisaO(?)(135):_  GD^CG  =  F(    (?)  (341). 

But  GD  x  CG  =  area  G'D'IH  (?).    F£2  =  j£C2=area  S.  (Explain.) 

.•.O6?'D'lH=o=-S(Ax.  6).   AlsoG'l>'+  G'C'  =  CD(?).    Q.E.D. 


S 

\F>-  -... 
\  X 

H      F(                                             1         ^ 

r                 X-°                U 

\             \ 

V                       E 

v  v 

E'                                                 G1 

II.  Given :  Square  S  and  line  CD. 

Required  :  To  construct  a  O  =c=  s ;  base  —  altitude  =  CD. 


BOOK  IV 


211 


Construction  :  On  CD  as  diameter,  describe  a  0,  o.  At  C 
erect  CEA.  to  CD  and  =AB.  Draw  EFOG  meeting  O  at  F  and 
G.  Take  E'G'^EG  and  draw  XY  II  to  E'G'  at  a  distance 
from  it  =  EF.  On  XY  take  JJI  =  #G.  Draw  HE1  and  IG'. 

Statement:  EH  E'G'  ill  ^  s  and  base  —  alt.  =  CD.          Q.E.F. 

Proof  :  #C  is  tangent  to  O  O  (?).    .'.  EG  •  EF^EC2^?)  (333). 

EG  •  EF=  area  O  E'G'IH  (?),  and  #c2  =  Jfi2  =  area  S  (?). 

FG  =  CD(?.  Q.E.D. 


412.  PROBLEM.    To  find  two  lines  whose  product  is  given  : 
I.  If  their  sum  is  also  given.  |  [The  game  ag  m  -j 

II.  If  their  difference  is  also  given,  j 

413.  PROBLEM.  To  construct  a  square  having  a  given  ratio  to  a 
given  square. 

O 


R 

/            Q/                 X*    \ 

/    E--'''                           \F\ 

S 

/      *r.-—                         ""\    \ 

/       ."""                                                                            *"•      • 

AL---''          m                       n         '•••         V 

X 

O 

A                                B                     C 

n....  , 

Given  :    Square  12,  and  lines  ra  and  n. 

Required  :    To  construct  a  square  such  that, 

The  square  E  :  the  unknown  square  =  m  :  n. 

Construction  :  On  an  indefinite  line  AY  take  AB  =  ra,  and 
BC  =n.  On  AC  as  diameter  describe  a  semicircle.  At  B 
erect  BD±.  to  AC,  meeting  arc  at  _D.  Draw  AD  and  DC.  On 
AD  take  DE=a,  and  draw  ^.F  ||  to  AC,  meeting  DC  at  F. 
Using  DF  =x,  as  a  side,  construct  square  S. 

Statement  :    R  :  S  =  m  :  n.  Q.E.F. 

Proof  :  Z  ^DC  is  a  rt.Z(?).    /.  ^D2  :  Z)C'2  =  m  :  w  (?)  (395). 


DC 


a2  = 


=  S  (?).     /.  I?  :  ,s  =  m  :  n  (Ax.  6). 


212 


PLANE   GEOMETRY 


414.   PROBLEM.  To  construct  a  polygon  similar  to  a  given  polygon 
and  having  a  given  ratio  to  it. 

D 


Given:    (?).     Required:    (?).    Construction  and  Statement 
are  the  same  as  in  413. 

Proof:    Z  ^DCisart.  Z  (?);     /.  AJ?  :  DC2  =  m  :  n  (?)(395). 

ff=^5(?);  .-.£!?  =  45-  =  ™.  (Explain.)    -  =  ^  (?)  (390). 

.'.  R  :  S  =  m  :  n  (Ax.  1).  Q.E.D. 

415.    PROBLEM.  To  construct  a  polygon  similar  to  one  given  poly- 
gon and  equivalent  to  another. 


A  B  a  —f 

Given  :    Polygons  R  and  S.    Required  :    (?). 
Construction  :  Construct  squares  R'  oR,  and  sf=o=8  (by  410). 

Find  a  fourth   proportional  to  a,  6,  and  AB.     This   is  CD. 

Upon  CD,  homologous  to  AB,  construct  T  similar  to  R. 
Statement :    T  ^  s.  Q.E.F. 


Proof  :  -  = 
T 


.-.2=?-    (Ax.  6).     .-.T^s     (Ax.  3). 


(?)  (390).  ^=^  (Const.).  .'.^==; 
b      CD  b2      CD' 


=c=/s.  (Explain.) 
Q.E.D. 


BOOK   IV  213 

ORIGINAL   CONSTRUCTIONS 

1.  To  construct  a  square  equivalent  to  a  given  right  triangle. 

2.  To  construct  a  right  triangle  equivalent  to  a  given  square. 

3.  To  construct  a  right  triangle  equivalent  to  a  given  parallelogram. 

4.  To  construct  a  square  equivalent  to  the  sum  of  two  given  right 
triangles. 

5.  To  construct  a  square  equivalent  to  the  difference  of  two  given 
right  triangles. 

6.  To  construct  a  square  equivalent  to  the  sum  of  two  given  paral- 
lelograms. 

7.  To  construct  a  square  equivalent  to  the  difference  of  two  given 
parallelograms. 

8.  To  construct  a  square  equivalent  to  the  sum  of  several  given  right 
triangles. 

9.  To  construct  a  square  equivalent  to  the  sum  of  several  given  paral- 
lelograms. 

10.  To  construct  a  square  equivalent  to  the  sum  of  several  given 
triangles.  • 

11.  To  construct  a  square  equivalent  to  the  sum   of  several  given 
polygons. 

12.  To  construct  a  square  equivalent  to  the  difference  of  two  given 
polygons. 

13.  To  construct  a  square  equivalent  to  three  times  a  given  square. 
To  construct  a  square  equivalent  to  seven  times  a  given  square. 

14.  To  construct  a  right  triangle  equivalent  to  the  sum  of  several 
given  triangles. 

15.  To  construct  a  right  triangle  equivalent  to  the  difference  of  any 
two  given  triangles;  of  any  two  given  parallelograms. 

16.  To  construct  a  square  equivalent  to  a  given  trapezoid  ;  equiva- 
lent to  a  given  trapezium. 

17.  To  construct  a  square  equivalent  to  a  given  hexagon. 

18.  To  construct    a  rectangle   equivalent  to   a  given  triangle,  hav- 
ing given  its  perimeter. 

19.  To  construct  an  isosceles  right  triangle  equivalent  to  a  given 
triangle. 

20.  To  construct  a  square  equivalent  to  a  given  rhombus. 

21.  To  construct  a  rectangle  equivalent  to  a  given   trapezium,  and 
having  its  perimeter  given. 


214  PLANE   GEOMETRY 

22.  To  find  a  line  whose  length  shall  be  \/2  units.     [See  402.] 

23.  To  find  a  line  whose  length  shall  be   V3  units. 

24.  To  find  a  line  whose  length  shall  be   v/11  units. 

25.  To  find  a  line  whose  length  shall  be  \/7  units. 

26.  To  find  a  line  whose  length  shall  be  VlO  units. 

27.  To  construct  a  square  which  shall  be  f  of  a  given  square. 

28.  To  construct  a  square  which  shall  be  f  of  a  given  square. 

29.  To  construct  a  polygon  which  shall  be  f  of  a  given  polygon,  and 
similar  to  it. 

30.  To  construct  a  square  which  shall  have  to  a  given  square  the 
ratio  V3  :  4.     If  the  given  ratio  is  4  :  V3. 

31.  To  draw  through  a  given  point,  within  a  parallelogram,  a  line 
which  shall  bisect  the  parallelogram. 

32.  To  construct  a  rectangle  equivalent  to  a  given  trapezoid,  and 
having  given  the  difference  of  its  base  and  altitude. 

33.  To  construct  a  triangle  similar  to  two  given  similar  triangles  and 
equivalent  to  their  sum. 

34.  To  construct  a  triangle  similar  to  a  given  triangle  and  equivalent 
to  a  given  square.     [See  415.] 

35.  To  construct  a  triangle  similar  to  a  given  triangle  and  equivalent 
to  a  given  parallelogram. 

36.  To  construct  a  square  having  twice  the  area  of  a  given  square. 
[Two  methods.] 

37.  To  construct  a  square  having  3|  times  the  area  of  a  given  square. 

38.  To  construct  an  isosceles  triangle  equivalent  to  a  given  triangle 
and  upon  the  same  base. 

39.  To  construct  a  triangle  equivalent  to   a  given  triangle,  having 
the  same  base,  and  also  having  a  given  angle  adjoining  this  base. 

40.  To   construct   a  parallelogram  equivalent    to  a  given  parallelo- 
gram, having  the  same  base   and  also  having  a  given  angle  adjoining 
the  base. 

41.  To  draw  a  line  that  shall  be  perpendicular  to  the  bases  of  a 
parallelogram  and  that  shall  bisect  the  parallelogram. 

42.  To  construct  an  equilateral   triangle  equivalent   to  a  given  tri- 
angle.    [See  415.] 

43.  To  trisect   (divide  into  three  equivalent  parts)  the   area  of  a 
triangle,  by  lines  drawn  from  one  vertex. 


BOOK   IV  215 

44.  To  construct  a  square  equivalent  to  £  of  a  given  pentagon. 

45.  To    construct    an    isosceles    trapezoid    equivalent    to    a    given 
trapezoid. 

46.  To  construct  an  equilateral  triangle  equivalent  to  the  sum  of  two 
given  equilateral  triangles. 

47.  To  construct  an  equilateral  triangle  equivalent  to  the  difference 
of  two  given  equilateral  triangles. 

48.  To  construct    upon    a    given   base    a    rectangle    that    shall   be 
equivalent  to  a  given  rectangle. 

Analysis :  Let  us  call  the  unknown  altitude  x. 
Then  b  .  h  =  b'  •  x  (V).  Hence,  b':b  =  h:x  (?) . 

That  is,  the  unknown  altitude  is  a  fourth  pro- 
portional to  the  given  base,  the  base  of  the  given  rec- 
tangle, and  the  altitude  of  the  given  rectangle. 

Construction:  Find  a  fourth  proportional,  x,  to  &',  b  and  h.     Con- 
struct a  rectangle  having  base  =  b'  and  alt.  =  x. 

Statement:  This  rectangle,  B**A. 


B 

Proof:  b':b  =  h:x  (Const.).   .-.  b'x  =  bh  (?).    But    ' 

b'x  =  the  area  of  B  (?).  Etc. 

49.  To  construct  a  rectangle  that  shall  have  a  given  altitude  and  be 
equivalent  to  a  given  rectangle. 

50.  To  construct  a  triangle  upon  a  given  base  that  shall  be  equiva- 
lent to  a  given  triangle. 

51.  To  construct  a  triangle  that  shall  have  a  given  altitude  and  be 
equivalent  to  a  given  triangle. 

52.  To  construct  a  rectangle  that  shall  have  a  given  base,  and  shall 
be  equivalent  to  a  given  triangle. 

53.  To  construct  a  triangle  that  shall  have  a   given  base,  and  be 
equivalent  to  a  given  rectangle. 

54.  To  construct  a  triangle  that   shall   have  a  given  base  and  be 
equivalent  to  a  given  polygon. 

55.  Construct  the  problems  49,  50,  51,  53,  54  if  the  first  noun  in  each 
problem  is  the  word  "  parallelogram." 

56.  To  construct  upon  a  given  hypotenuse,  a  right  triangle  equivalent 
to  a  given  triangle. 

57.  To  construct  upon  a  given  hypotenuse,  a  right  triangle  equivalent 
to  a  given  square. 


216  PLANE   GEOMETRY 

58.  To  construct  a  triangle  which  shall  have  a  given  base,  a  given 
adjoining  angle,  and  be  equivalent  to  a  given  triangle.     Another,  equiva- 
lent to  a  given  square.     Another,  equivalent  to  a  given  polygon. 

59.  To  construct  a  parallelogram  which  shall  have  a  given  base,  a 
given   adjoining   angle,   and  be  equivalent  to   a  given  parallelogram. 
Another,  equivalent  to  a  given  triangle;  to  a  given  polygon. 

60.  To  construct  a  line,  DE,  from  D  in  AB  of  triangle  ABC,  so  that 
DE  bisects  the  triangle.  A 

Analysis:  After  DE  is  drawn,  ^ABC='2AADE 
(Hyp.).  Rnt&ABC:  &  ADE=AB  •  AC :  AD  >  AE  (?). 
Hence,  AB  .  AC  =  2  (AD  .  AE)  (Ax.  6). 

/.  2  AD :  AB  =  A  C :  x  (?).  Thus  x,  (that  is,  AE)  is 
a  fourth  proportional  to  three  given  lines. 

61.  To  draw  a  line  meeting  two  sides  of  a  triangle  and  forming  an 
isosceles  triangle  equivalent  to  the  given  triangle. 

Analysis:  Suppose  AX  a  leg  of  isosceles  A. 
/ .  A  A  B  C :  A  A  XX'  =  A  B  .  A  C :  A  X  .  A  X' . 
But  the  A  are  equivalent  and  A X  =  A  X'   (Hyp.). 
Hence,  AB  •  A  C  =  A  X2.    .'.  AX  is  a  mean  proper-     B  \c 

tional  between  AB  and  AC. 

62.  To  draw  a  line  parallel  to  the  base  of  a  triangle  which  shall 
bisect  the  triangle.     [See  389  and  use  414.] 

63.  To   draw    a   line    meeting   two  sides   of  a  triangle  forming   an 
isosceles  triangle  equivalent  to  half  the  given  triangle. 

64.  To  draw  a  line  parallel  to  the  base  of  a  triangle  forming  a  tri- 
angle equivalent  to  one  third  the  original  triangle. 

65.  To  draw  a  line  parallel  to  the  base  of  a  O 
trapezoid  so  that  the  area  is  bisected.                                             /'\ 

Analysis:   A  OXX'  =*%   (A  OAD  +  A  OB  C)  /     \ 

and  is  similar  to  A  OB  C.  «/ \c 

Construction:  [Use  408, 402, 415.]  /  ^ 

66.  To  construct  two  lines  parallel  to  the  base       / \ 

of  a  triangle,  that  shall  trisect  the  area  of  the  A  D 

triangle. 

67.  To  construct  a  triangle  having  given  its  angles  and  its  area. 
Analysis :  The  required  A  is  similar  to  any  A  containing  the  given 

A.     The  given  area  may  be  a  square.     This  reduces  the  problem  to  415. 

68.  To  find  two  straight  lines  in  the  ratio  of  two  given  polygons. 


BOOK  V 


REGULAR   POLYGONS.     CIRCLES 

416.  A  regular  polygon  is  a  polygon  which  is  equilateral 
and  equiangular. 

417.  THEOREM.  An  equilateral  polygon  inscribed  in  a  circle  is  regular. 

Given  :    AG,  an  equilateral 
inscribed  polygon. 

To  Prove  :    AG  is  regular. 

Proof  :    Z  A  is  measured  by 
1  arc  BGL  (?). 

Also  Z  B  is    measured    by 
Jarc  CHA  (?),  etc. 

Subtended  arcs,  AB,  BC,  CD, 
etc.,  are  all  =  (?). 

Hence,  arcs  BGL,  CHA,  DIB, 
etc.,  are  all  =  (Ax.  2). 

/.  /.A  =  ZB  =  Zc=  etc.    (?). 

That  is,  the  polygon  is  equiangular. 

Therefore,  the  polygon  is  regular  (?)  (416).  Q.E.D. 

418.  THEOREM.   If  the  circumference  of  a  circle  be  divided  into  any 
number  of  equal  arcs,  and  the  chords  of  these  arcs  be  drawn,  they  will 
form  an  inscribed  regular  polygon. 

Proof:    Chords  AB,  B(\  CD,  etc.  are  all  =  (?). 

Therefore,  the  polygon  is  regular  (?).  Q.E.D. 


Ex.  1.    What  is  the  usual  name  of  a  regular  3-gon  ?   of   a   regular 
4-gon?    Is  an  equiangular  inscribed  polygon  necessarily  regular? 

Ex.  2.   In  the  figure  of  417,  how  many  degrees  are  there  in  each  of  th^ 
arcs,  AB,  BC,  etc.?     How  many  degrees  are  there  in  each  angle? 

217 


218 


PLANE   GEOMETRY 


419.  THEOREM.   If  the  circumference  of  a  circle  be  divided  into  any 
number  of  equal  parts,  and  tangents  be  drawn,  at  the  several  points  of 
division,  they  will  form  a  circumscribed  regular  polygon. 

Given  :  (?).  To  Prove :  (?). 

Proof  :    Draw    chords   AB, 
BC,  CD,  etc. 

In  A  ABH,  BCI,  CDJ,  etc., 
AB  =  BC  =  CD  =  etc.  (?). 

Z  HAB  =  Z  HBA  =  Z IBC  = 
Z.ICB  =  Z  JCD  =  etc.  (?). 

.*.  these  A  are  isosceles  (?) 
(120),  and  =  (?). 
.-.  Z  H  =  Z  1  =  Z  j  =etc.  (?). 
That  is,  polygon  GJ  is  equi- 
angular. 

Also,  AH  =  HB  =  BI  =  1C 
=  CJ  =etc.  (?),  and  HI  =  IJ  =  JK  =  etc.  (?)  (Ax.  3). 

That  is,  polygon  £Jis  equilateral ;  .*.  it  is  regular  (?).  Q.E.D. 

420.  THEOREM.     If    the    cir- 
cumference of  a  circle  be  divided 
into  any   number  of  equal  parts 
and  tangents  be  drawn  at  their 
midpoints,  they  will  form  a  cir- 
cumscribed regular  polygon. 

Given:  (?).  To  Prove:  (?).    G 

Proof  :    Arcs   AB,   BC,   CD, 
etc.  are  all   =  (?). 

Also,  arcs  A  I,  IB,  BK,  KG,   R 
CM,  etc.  are  all  =  (?)  (Ax.  3). 

.*.  arcs  IK,  KM,  MO,  etc.  are 
all  =  (?)  (Ax.  3). 

Therefore,  the  polygon  is  regular  (?)  (419).  Q.E.D. 

421.  THEOREM    If  chords  be  drawn  joining  the  alternate  vertices 
of  an  inscribed  regular  polygon  (having  an  even  number  of  sides), 
another  inscribed  regular  polygon  will  be  formed.     (See  417.) 


BOOK   V 


219 


422.  THEOREM.    If  the   ver- 
tices   of    an    inscribed    regular 
polygon  be  joined  to   the  mid- 
points of  the  arcs  subtended  by 
the  sides,  another  inscribed  regu- 
lar polygon  will  be  formed  (hav- 
ing double  the  number  of  sides). 

423.  THEOREM.    If   tangents 
be  drawn  at  the  midpoints  of  the 
arcs  between  adjacent  points  of 
contact  of  the  sides  of  a  circum- 
scribed regular  polygon,  another  circumscribed  regular  polygon  will 
be  formed  having  double  the  number  of  sides  (?). 

424.  THEOREM.    The  perimeter  of  an  inscribed  regular  polygon  is 
less  than  the  perimeter  of  an  inscribed  regular  polygon  having  twice 
as  many  sides,  and  the  perimeter  of  a  circumscribed  regular  polygon 
is  greater  than  the  perimeter  of  a  circumscribed  regular   polygon 
having  twice  as  many  sides.     (Ax.  12.) 

425.  THEOREM.    Two    regular 
polygons  having  the  same  number 
of  sides  are  similar. 

Given  :  Regular  w-gons  AD 
and  A'D'. 

To  Prove  :  They  are  simi- 
lar. 


Proof  : 


?)  (164). 


-2)  180° 
n 


Similarly,  Z  B  =  Z  Bf,  Z  C==  Z  Cf,  etc. 
That  is,  these  polygons  are  mutually  equiangular. 
Also,  ^17?  =  BC  =  CD=  etc.;  A'B'  =  B'C'  =  c'z)'  = 
.-.  AB  :  A'B'  =  BC:  Bfc'  =  CD  :  CfDf=  etc.  (Ax.  3). 
That  is,  the  homologous  sides  are  proportional. 
Therefore,  the  polygons  are  similar  (?). 


Q.E.D. 


220 


PLANK    (JKOMKTKY 


426.  THEOREM.   A  circle  can  be  circumscribed  about,  and  a  circle  can 
be  inscribed  in,  any  regular  polygon. 

Given :  Regular  polygon 
ABCDEF. 

To  Prove :  I.  A  circle  can 
be  circumscribed  about  the 
polygon. 

II.  A  circle  can  be  in- 
scribed in  the  polygon. 

Proof:    I.     Through  three 
consecutive    vertices,    A,    B, 
and   C,  describe  a  circumfer- 
ence, whose  center  is  O.    Draw  radii  OA,  OJ5,  OC,  and  draw 
line  OD. 

In  A  AOB  and  COD,  AB  =  CD  (?)  ;   BO  =  CO  (?). 

Now  ^ABC  =  ZBCD  (?)  (416).     /Lose  =  Z  OCB  (?). 
Subtracting,  Z  ABO  =  Z.  OCD  (Ax.  2). 

.'.A  AOB  =  A  COD  (?).      .'.AO  =  OD  (?). 

Hence,  the  arc  passes  through  D,  and  in  like  manner  it 
may  be  proven  that  it  passes  through  E  and  F. 

That  is,  a  circle  can  be  circumscribed  about  the  polygon. 

II.    AB,  BC,  CD,   DE,  etc.   are  =  chords  (?)  (416). 

Therefore  they  are  equally  distant  from  the  center  (?). 

That  is,  a  circle  described,  using  O  as  a  center  and  OM  as  a 
radius,  will  touch  every  side  of  the  polygon. 

Hence  a  circle  can  be  inscribed.     (Def.  234.)  Q.E.D. 

427.  The  radius  of  a  regular  polygon  is  the  radius  of  the 
circumscribed  circle.     The  radius  of  the  inscribed  circle  is 
called  the  apothem.     The  center  of  a  regular  polygon  is  the 
common  center  of  the  circumscribed  and  inscribed  circles. 

428.  The  central  angle  of  a  regular  polygon  is  the  angle 
included  between  two  radii  drawn  to  the  ends  of  a  side. 


429.   THEOREM.   Each  central  angle  of  a  regular  w-gon  = 


BOOK   V 


221 


430.  THEOREM.   Each  exterior  angle  of  a  regular  /7-gon  =  * —  (?)• 

431.  THEOREM.  The  radius  drawn  to  any  vertex  of  a  regular  poly- 
gon bisects  the  angle  at  the  vertex.     (See  80.) 

432.  THEOREM.  The  central  angles  of  regular  polygons  having  the 
same  number  of  sides  are  equal.    (See  429.) 

433.  THEOREM.   If  radii  be  drawn  to  all  the  vertices  of  a  circum- 
scribed regular  polygon,  and  chords  be  drawn  connecting  the  points  of 
intersection  of  these  lines  with  the  circumference,  an  inscribed  regular 
polygon  of  the  same  number  of  sides  will  be  formed  and  the  sides  of  the 
two  polygons  will  be  respectively  parallel. 

Given:  (?).    To  Prove:  (?). 

Proof :  Central  A  at 
O  are  all  =  (?). 

.-.the  intercepted  arcs 
are  all  =  (?). 

.-.  the      chords      A1 
B'c',  etc.  are  all=  (?). 

.-.  the  inscribed  poly- 
gon is  regular  (?). 

Also,  A  AOB,  A'OB', 
etc.  are  isosceles  (?). 

If  a  line  OX  be  drawn 
from  O,  bisecting  Z  AOB, 
it  is  _L  to  AB  and  to  A'B'  (?)  (57). 

.'.  AB  is  ||  to  A'B'  (?).  Q.E.D. 


Ex.  1.  How  many  degrees  are  there  in  the  angle,  in  the  central  angle, 
and  in  the  exterior  angle  of  a  regular  hexagon?  a  regular  decagon? 
a  regular  15-gon  ? 

Ex.  2.    In  the  figure  of  433,  prove, 

(rt)  Triangle  A  OB  similar  to  triangle  A'OB'. 

(it)  Triangle  XOB  similar  to  triangle  X'OB'. 


222 


PLANE   GEOMETRY 


434.  THEOREM.  The  perimeters  of  two  regular  polygons  having 
the  same  number  of  sides  are  to  each  other  as  their  radii  and  also  as 
their  apothems. 


A'  B1 

Given  :  Regular  w-gons,  EC  whose  perimeter  is  P,  radius  JR, 
apothem  r ;  and  EfCr  whose  perimeter  is  P',  radius  R1, 
apothem  r'. 

To  Prove  :   P  :  P'  =  R  :  Bf  =  r  :  r' . 

Proof:   Draw  radii  OB  and  O'B'.      In  ,A  AOB  and  A!OrBf, 


Hence,  4^-,  =  4^7  (Ax.  3). 

A'Of       B'Of 
.-.  A  AOB  is  similar  to  A  A!OfBf  (?)  (317). 

A  7?  7?          T 

.'.  —f—j  =  —f  =  —  (?).     Also,  the  polygons  are  similar  (?). 


435.  THEOREM.  The  areas  of  two  regular  polygons  having  the 
same  number  of  sides  are  to  each  other  as  the  squares  of  their 
radii  and  also  as  the  squares  of  their  apothems. 

Proof  :   If  K  and  Kf  denote  their  areas,  we  have  : 


(?)(390).          But 


TJ  K      R2       r2  ,  A       .<>. 

Hence,  _  =  _  =  _( Ax.  1). 


--          (Explain.) 


Q.E.D. 


ROOK   V 


223 


436.   THEOREM.    The  area  of  a  regular  polygon  is  equal  to  half  the 
product  of  the  perimeter  by  the  apothem. 

Given:   (?). 

To  Prove:  (?). 

Proof  :  Draw  radii  to  all  the 
vertices,  forming  several  isos- 
celes triangles. 

Area  of  A  AOB  =  \AH  -  r 

•       ,?, 


Area  of  A  BOC  =  J  BC  •  r 
Area  of  A  COD  =  \  CD  •  r 

etc.,  etc. 
Area  of  polygon  =  1  (A  B  -f  BC  +  CD  +  etc.)  •  r  (?), 


or, 


=  ^  P-  r  (Ax.  6). 


Q.E.D. 


437.  THEOREM.  If  the  number  of  sides  of  an  inscribed  regular  poly- 
gon be  increased  indefinitely,  the  apothem  will  approach  the  radius  as 
a  limit.  g^*- ""^p 

Given  :  N-gon  FC  inscribed  in 
O  O ;   apothem  =  r  ;  radius  =  if. 

To  Prove:  That  as  the 
number  of  sides  is  indefi- 
nitely increased,  r  approaches 
R  as  a  limit. 

Proof:  In  the  A  AOK, 

or  R  -r<AK  (?). 

Now  as  the  number  of  sides  of  the  polygon  is  indefinitely 
increased,  AB  will  be  indefinitely  decreased. 

Hence,  £  AB,  or  AK,  will  approach  zero  as  a  limit. 

.'.  R  —  r  will  approach  zero  (because  R  —  r  <  AK). 

That  is,  r  will  approach  R  as  a  limit  (240).  Q.E.D. 

NOTE.  It  is  evident  that  if  the  difference  between  two  variables 
approaches  zero,  either 

(1)  one  is  approaching  the  other  as  a  limit ;  or 

(2)  both  are  approaching  some  third  quantity  as  their  limit. 


224  PLANK   GEOMETRY 

Ex.  1.  The  apothem  of  a  regular  polygon  perpendicular  to  a  side 
bisects  the  side  and  the  central  angle  of  the  polygon. 

Ex.  2.  Are  all  pairs  of  squares  similar?  Why?  Are  all  pairs  of 
rhombuses  similar?  Why?  Is  a  rhombus  a  regular  polygon?  Is  a, 
rectangle  a  regular  polygon  ? 

Ex.    3.    What  is  the  area  of  a  square  whose  perimeter  is  40  inches  ? 

Ex.  4.  What  is  the  area  of  a  regular  hexagon  whose  side  is  8  and 
apothem  is  4  V^i? 

Ex.    5.   What   is  the   area   of  a  regular   dodecagon    whose  side    is 

10  ^2  -  V3  and  whose  apothem  is  5  ^2  +  Vl  ? 

Ex.  6.  Prove  that  the  lines  joining  the  midpoints  of  the  sides  of  a 
regular  polygon,  in  order,  form  a  regular  polygon  of  the  same  number 
of  sides. 

Ex.  7.  Prove  that  a  polygon  is  regular  if  the  inscribed  and  circum- 
scribed circles  are  concentric. 

Ex.  8.  Draw  a  figure  showing  an  inscribed  equiangular  polygon  that 
is  not  regular. 

Ex.  9.  Draw  a  figure  showing  a  circumscribed  equilateral  polygon 
that  is  not  regular. 

Ex.  10.  Prove  that  the  circumference  of  a  circle  is  greater  than 
the  perimeter  of  any  inscribed  polygon. 


438.  The  theorems  of  439  and  440  are  considered  so  evident,  and 
rigorous  proofs  are  so  difficult  for  young  students  to  comprehend  (like 
the  demonstrations  for  many  fundamental  principles  in  mathematics) 
that  it  is  advisable  to  omit  the  profound  demonstrations  and  insert  only 
simple  explanations. 

439.  THEOREM.   The  circumference  of  a  circle  is  less  than  the  per- 
imeter of  any  circumscribed  polygon. 

By  drawing  tangents  at.  the  midpoints  of  the  included 
arcs  another  circumscribed  polygon  is  formed;  the  perim- 
eter of  this  polygon  is  less  than  the  perimeter  of  the  given 
polygon  (?).  This  can  be  continued  indefinitely,  decreas- 
ing the  perimeter  of  the  polygons.  Hence,  there  can  be  no 
circumscribed  polygon  whose  perimeter  can  be  the  least  of 
all  such  polygons  ;  because,  by  increasing  the  number  of  sides, 
the  perimeter  is  lessened.  Hence,  the  circumference  must 
be  less  than  the  perimeter  of  any  circumscribed  polygon. 


HOOK    V 


440.  THEOREM.  If  the  number  of  sides  of  an  inscribed  regular  poly 
gon  and  of  a  circumscribed  regular  polygon  be  indefinitely  increased, 
I.  The  perimeter  of  each  polygon  will  approach  the  circumference 
of  the  circle  as  a  limit. 

II.  The  area  of  each  polygon  will  approach  the  area  of  the  circle 
as  a  limit. 

Given:  A  circle  O,  whose  cir- 
cumference is  C  and  area  is  8  ; 
AB  and  A'B',  sides  of  regular  cir- 
cumscribed and  inscribed  poly- 
gons, having  the  same  number  of 
sides ;  P  and  p',  their  perimeters ; 
K  and  K1,  their  areas. 

To  Prove  :  That  if  the  number 
of  sides  be  indefinitely  increased  : 

I.  P  will  approach  C  and  P'  will  approach  C  as  limit. 
II.  K  will  approach  8  and  K'  will  approach  8  as  limit. 
Proof:  I.  The  polygons  are  similar  (?)  (425). 

.-.  —  =  — (?)  (434).     Now,  if  the   number   of   sides   of 

these   polygons  be  indefinitely  increased,  OD  will  approach 
OE  (?)  (437). 

Hence, — will  approach  1.    That  is,—  will  approach  1,  or  P 

and  Pr  will  approach  equality  ;  that  is,  they  will  approach  the 
same  constant  as  a  limit. 

But  P  >  C    and  C  >  P1  and  C  is  constant. 

Hence,  P  will  approach  C  and  p'  will  approach  C.      Q.E.D. 


ii.  f=J|<?X486)- 


If  the  number  of  sides  of  these 


polygons  be  indefinitely  increased,  OD2  will  approach  O.E2,  and 


OE 


thus  — -will  approach  unity. 

OD 
(The  argument  continues  the  same  as  in  I.) 


BOBBINS'  PLANE  GEOM. — 15 


226 


PLANE   GEOMETRY 


441.  THEOREM.    The   circumferences  of  two  circles  are   to  each 
other  as  their  radii. 

Given  :  Two  CD  whose  radii 
are  R  and  R*  and  circumfer- 
ences, C  and  cr  respectively. 

To  Prove  :   C  :  C1  =  R  :  Rf. 

Proof:  Circumscribe  regu- 
lar polygons  (having  the 

same  number  of  sides)  about  these  (D  and  let  P  and  p' 
denote  their  perimeters.  Then,  P  :  Pr  =  R  :  Rf  (?)  (434). 
Hence,  P  •  Rr  =  P'  •  R  (?).  Now  suppose  the  number  of 
sides  of  these  polygons  to  be  indefinitely  increased, 

P   will  approach  c  (?)  (440). 
Pf  will  approach  c'  (?). 
.*.  P  •  Rr  will  approach  c  -  -B', 
and  P1  -  R  will  approach  cr  •  R. 

Hence,  C  -  Rf  =  c'  -  R  (?)  (242). 

Therefore          C  :  Cf  =  R  :  Rf  (?)  (291).  Q.E.D. 

442.  THEOREM.    The  ratio  of  any  circumference  to  its  diameter  is 
constant  for  all  circles.     That  is,  any  circumference  divided  by  its 
diameter  is  the  same  as  any  other  circumference  divided  by  its  diam- 
eter. 

Proof:  -  =  ^-  (?)  (441).   But    *.  =  §£-«-£(?). 

c'     R'  R'     %R'     D' 

.--!  =  !,  (A*.  1). 
Hence,  £«..£..(?)  (292).  That  is,  -  =  constant.        Q.E.D. 

443.  Definition  of  TT  (pi) .      The  constant  ratio  of  a  circum- 

c1 

ference  to  its  diameter  is  called  TT.      That  is,  —  ==  TT. 

D 


BOOK    V 


227 


The  numerical  value  of  TT  =  3.141592  =  3^,  approximately. 
(This  is  determined  in  470.) 

444.   FORMULA.    Let  C  =  circumference  and  R  =  radius. 
Then,  ^  =  TT  (443).      .-.  C  =  27rR   (Ax.  3). 


445.   THEOREM.    The  area  of  a  circle  is  equal  to  half  the  product  of 
its  circumference  by  its  radius. 

Given:  O  whose  circumfer- 
ence =  C',  area  =6',  radius  =  R. 

To  Prove  :  s  =  I  c  •  R. 

Proof  :  Circumscribe  a  reg- 
ular polygon  about  the  circle  ; 
denote  its  area  by  K  and  per- 
imeter by  P. 

Nowir=ip..R(?)(436). 

Suppose  the  number  of 
sides  of  the  polygon  be  in- 
definitely increased. 

K  will  approach  s,  and  P  will  approach  C  (?). 

\  P  •  R  will  approach  \  C  •  R  as  a  limit  (?). 
Hence,  s  =  i  C  -  R  (?)  (242). 


Q.E.D. 


446.  FORMULA.    Let  S  =  area  of  O,  C  =  its  circumference,  and 
R  =  its  radius.     Then,  S  =  J  C  •  R    (445). 

Now  C=  2  TT  £  (444).     Substituting,  S  =  £  (2  TTR)  fi. 

.'.     S=   7TJR2. 


Ex.  1.  Could  445  be  proven  by  inscribing  a  regular  polygon?    Why? 

Ex.  2.  The  radius  of  a  circle  is  40.    Find  the  circumference  and  area. 

Ex.  3.  The  diameter  of  a  circle  is  25.    Find  the  circumference  and  area. 

Ex.  4.  Prove  that  the  area  of  a  circle  equals  .7854  D-. 


228  PLANE   GEOMETRY 

447.   THEOREM.    The  areas  of  two  circles  are  to  each  other  as  the 
squares  of  their  radii,  and  as  the  squares  of  their  diameters. 

To  Prove:    8  :  S1  =  R2:  Rt2  =  D2:Df2. 
Proof : 


.'.  S:S'=R2:Rf2  =  D2:  D12  (Ax.  1).  Q.E.D. 

448.  THEOREM.   The  area  of  a  sector  is  the  same  part  of  the  circle 
as  its  central  angle  is  of  360°.     (Ax.  1.) 

449.  FORMULA.   An  arc  :  circum.  =  central Z  :  360°  (244). 

. .  arc  :  2  TTR  =  Z  :  360°. 

NOTE.    If  any  two  of  the  three  quantities,  arc,  R,  Z,  are  known,  the 
remaining  one  can  be  found  by  this  proportion. 

450.  FORMULA.    Sector  :  area  of  O  =  central  Z:  360°  (448). 

. .  sector  :  TrJB2  =  Z  :  360°. 

NOTE.    If  any  two  of  the  three  quantities,  sector,  R,  Z,  are  known,  the 
remaining  one  can  be  found  by  this  proportion. 

451.  FORMULA.    Sector  :  area  of  0  =  arc  :  circum.  (Ax.  1). 
.-.  sector  :  TT  R2  =  arc  :  2  TTR    (Ax.  6). 

. .  sector  =  %R  •  arc  (290). 

452.  FORMULA.  Area  of  a  segment  of  a  circle  =  area  of  the 
sector  minus*  area  of  an  isosceles  triangle. 

453.  Similar  arcs,  similar  sectors,  and 
similar  segments  are  those  which  corre- 
spond to  equal  central  angles,  in  unequal 
circles. 

Thus,  AB,  A'B',  A"B"  are  similar  arcs; 
AOB,  A' OB1,  and  A" OB"  are  similar  sec- 
tors ;  and  the  shaded  segments  are  simi- 
lar segments. 

*  If  the  segment  is  greater  than  a  semicircle,  the  area  of  the  triangle  should 
be  added. 


BOOK  V 


454.   THEOREM.    Similar  arcs  are  to  each  other  as  their  radii. 
Given  :   Arcs  whose  lengths  are  a  and  a',  radii  R  and  R1. 
To  Prove  :  a  :  a'  —  R  :  Rf. 


.•.a:a'  =  27rfl:27rtf'(292);  /.a  :  a'  =  R:  K1  . 


.E.D. 


455.   THEOREM.  Similar  sectors  are  to  each  other  as  the  squares  of 
their  radii. 

Given  :   Sectors  whose  areas  are  T  and  T',  radii  R  and  Br. 
To  Prove  :   T  :  T'  =  R2  :  R12. 


=  _?!_(?).     .-.  T:T'=R2:R'2.  (Explain.)     Q.E.D. 


456.   THEOREM.    Similar  segments  are  to  each  other  as  the  squares 
of  their  radii. 

Given:  (?).     To  Prove :  (?).  o 

Proof:  A  AOB  and  A'O'B'  are 
similar  (?)  (317). 
A  AOB         »2 


R 


x9x    ^QOQX 


and 


sector 


sector  O'A  C'B' 
sector  OACB 

sector  O'A'C'B' 
sector  OACB  — 


-  , 


'  sector  ofA'c'B'—AAfo'B'~  A  A'O'B 


_       ^  /295    Note 

f''~        ''' 


TT  segment  ABC         A  AOB         R    s\       a\ 

Hence, — =_  -=-     — —  =  — -  (Ax.  6).       ~-« 

segment  A'B'C'     A  A'O'B'     R'2  Q.E.D. 


230  PLANE   GEOMETRY 

ORIGINAL    EXERCISES    (THEOREMS) 

1.  The  central  angle  of  a  regular  polygon  is  the  supplement  of  the 
angle  of  the  polygon. 

2.  An  equiangular  polygon  inscribed  in. a  circle  is  regular  (if  the 
number  of  its  sides  is  odd). 

3.  An  equiangular  polygon  circumscribed  about  a  circle  is  regular. 
[Draw  radii  and  apothems.] 

4.  The  sides  of  a  circumscribed  regular  polygon  are  bisected  at  the 
points  of  contact. 

5.  The  diagonals  of  a  regular  pentagon  are  equal. 

6.  The  diagonals  drawn  from  any  vertex  of  a  regular  n-gon  divide 
the  angle  at  that  vertex  into  n-2  equal  parts. 

7.  If  a  regular  polygon  be  inscribed  in  a  circle  and  another  regular 
polygon  having  the  same  number  of  sides  be  circumscribed  about  it,  the 
radius  of  the  circle  will  be  a  mean  proportional  between  the  apothem  of  the 
inner  and  the  radius  of  the  outer  polygon. 

8.  The  area  of  the  square  inscribed  in  a  sector 
whose  central  angle  is  a  right  angle  is  equal  to  half 
the  square   of  the   radius. 

[Find  z2,  the  area  of  OEDC.~\  /* 


9.  The  apothem  of  an  equilateral  triangle  is  one    O 
third  the  altitude  of  the  triangle. 

10.  The  chord  which  bisects  a  radius  of  a  circle  at  right  angles  is  the 
side  of  the  inscribed  equilateral  triangle. 

[Prove  the  central  Z  subtended  is  120°.] 

11.  If  ABCDE  is  a  regular  pentagon,  and 
diagonals  A  C  and  BD  be  drawn,  meeting  at  O : 

(a)  AO  will  =  AB. 

(b)  A  0  will  be  ||  to  ED. 

(c)  ABOC  will  be  similar  to  A  BDC. 

(e)  A  C  will  be  divided  into  mean   and  ex- 
treme ratio  at  O.  A 

12.  The  altitude  of  an  equilateral  triangle  is  three  fourths  the  diameter 
of  the  circumscribed  circle. 


BOOK   V 


231 


13.  The  apothem  of  an  inscribed  regular  hexagon  equals  half  the  side 
of  an  inscribed  equilateral  triangle. 

14.  The   area  of  a  circle  is  four  times  the  area  of   another  circle 
described  upon  its  radius  as  a  diameter. 

15.  The  area  of  an  inscribed  square  is  half  the  area  of  the  circum- 
scribed square. 

16.  An  equilateral  polygon  circumscribed  about  a  circle  is  regular 
(if  the  number  of  its  sides  is  odd). 

17.  The  sum  of  the  circles  described  upon  the  legs  of  a  right  triangle 
as  diameters  is  equivalent  to  the  circle  described 

upon  the  hypotenuse  as  a  diameter. 

18.  A  circular  ring  (the  area  between  two  con- 
centric circles)  is  equivalent  to  the  circle  described 
upon  the  chord  of  the  larger  circle,  which  is  tan- 
gent to  the  less,  as  a  diameter. 

Proof:  Draw  radii  OB,  OC.     A  OBC  isrt.A(?); 
and  OC'2  -  OB*  =  BC*  (?).     Etc. 

19.  If  semicircles  be  described  upon  the  three 
sides  of  a  right  triangle  (on  the  same  side  of  the 
hypotenuse),  the  sum  of  the  two  crescents  thus 
formed  will  be  equivalentto  the  areaof  the  triangle. 


Proof: 


f  Entire  figure  =c=  },r  AB2  +  crescent  BDC  +  crescent  AEC  (?). 


\Entire  figure-  ITT  AC2  +  %7rBC2+  A  ABC    (?). 
Now  use  Ax.  1  ;  etc. 

20.  Show  that  the  theorem  of  No.  19  is  true  in  the  case  of  a  right  tri- 
angle whose  legs  are  18  and  24. 

21.  If  from  any  point  in  a  semicircumference  a  line  be  drawn  perpen- 
dicular to  the  diameter  and   semicircles  be  de- 

scribed on  the  two  segments  of  the  hypotenuse  as 
diameters,  the  area  of  the  surface  bounded  by 
these  three  semicircumferences  will  equal  the 
area  of  a  circle  whose  diameter  is  the  perpendicu- 
lar first  drawn. 

Pttrf 


:  Area  =  1  ,('!±!)2  -  4  ,(|)2  -  4  ,(|)2=  etc. 

22.    Show  that  the  theorem  of  No.  21  is  true  in  the  case  of  a  circle 
whose  diameter  AB  is  25  and  AD  is  5. 


232  PLANE   GEOMETRY 

23.  If  the  sides  of  a  circumscribed  regular  polygon  are  tangent  to  the 
circle  at  the  vertices  of  an  inscribed  regular  polygon,  each  vertex  of  the 
outer  lies  on  the  prolongation  of   the  apothems  of  the  inner  polygon, 
drawn  perpendicular  to  the  several  sides. 

24.  The  sum  of  the  perpendiculars  drawn  from  any  point  within  a 
regular  n-gon  to  the  several  sides  is  constant  [=  n  •  apothem]. 

Proof :  Draw  lines  from  the  point  to  all  vertices.     Use  436  and  378. 

25.  The  area  of  a  circumscribed  equilateral  triangle  is  four  times  the 
area  of  the  inscribed  equilateral  triangle. 

26.  If  a  point  be  taken  dividing  the  diameter  of  a  circle  into  two  parts 
and  circles  be  described  upon  these  parts  as  diameters,  the  sum  of  the  cir- 
cumferences of  these  two  circles  equals  the  circumference  of  the  original 
circle. 

27.  Show  that  the  theorem  of  No.  26  is  true  in  the  case  of  a  circle 
the  segments  of  whose  diameter  are  7  and  12. 

28.  The  area  of  an  inscribed  regular  octagon  is  equal  to  the  product  of 
the  diameter  by  the  side  of  the  inscribed  square. 

29.  If  squares  be  described  on  the  six  sides  of  a  regular  hexagon 
(externally),  the  twelve  exterior  vertices  of  these  squares  will  be  the 
vertices  of  a  regular  12-gon. 

30.  If  the  alternate  vertices  of  a  regular  hexagon  be  joined  by  draw- 
ing diagonals,  another  regular  hexagon  will  be  formed.     Also  its  area 
will  be  one  third  the  original  hexagon. 

31.  Show  that  the  theorem  of  No.  18  is  true  in  the  case  of  two  con- 
centric circles  whose  radii  are  34  and  16. 

32.  In  the  same  or  equal  circles  two  sectors  are  to  each  other  as  their 
central  angles. 

33.  If  the  diameter  of  a  circle  is  10  in.  and  a  point  be  taken  dividing 
the  diameter  into  segments  whose  lengths  are  4  in.  and  6  in.,  and  on  these 
segments  as  diameters  semicircumferences  be  described  on  opposite  sides 
of  the  diameter,  these  arcs  will  form  a  curved  line  which  will  divide  the 
original  circle  into  two  parts  in  the  ratio  of  2  :  3. 

34.  If  the  diameter  of  a  circle  is  d  and  a  point  be  taken  dividing  the 
diameter  into  segments  whose  lengths  are  a  and  d  —  a,  and  on  these  seg- 
ments as  diameters  semicircumferences  be  described  on  opposite  sides  of 
the  diameter,  these  arcs  will  form   a  curved  line  which  will  divide  the 
original  circle  into  two  parts  in  the  ratio  of  a  :  d  —  a. 


BOOK  V 


233 


CONSTRUCTIONS 

457.  PROBLEM.    To  inscribe  a  square  in  a  given  circle. 

Given :  The  circle  o.  Re- 
quired :  To  inscribe  a  square. 

Construction :  Draw  any  di- 
ameter, AB,  and  another  diame- 
ter, CD,  -L  to  AB.  Draw  AC,  BC, 
BD,  AD. 

Statement:  ACBD  is  an  in- 
scribed square.  Q.E.F. 

Proof:  ,4  at  Oare=  (?). 

.-.  arcs  AC,  CB,  etc.  are  =  (?). 

.'.  ACBD  is  an  inscribed  regular  polygon  (?)  (418). 

.'.  ABCD  is  a  square  (?).  Q.E.D. 

458.  PROBLEM.    To  inscribe  a  regular  hexagon  in  a  given  circle. 
Given:  (?).     Required:  (?). 

Construction :  Draw  any  ra- 
dius, AO.  At  A,  with  radius 
=  AO,  describe  arc  intersecting 
the  given  O  at  B.  Draw  AB. 

Statement :  AB  is  the  side  of 
an  inscribed  regular  hexagon. 

Proof :  Draw  BO.  A  ABO  is 
equilateral  (Const.). 

.-.A  JBOisequiangular(?)(56).     .-.  Z  AOB  =  60°(?)(H5). 

.•.arc  AB  =^  of  the  circumference  (J  of  360°). 

.'.  polygon   AD,  inscribed,   having   each  side  =  AB,  is  an 
inscribed  regular  hexagon  (?)  (418).  Q.E.D. 


Ex.   If  the  radius  of  a  circle  is  7  in.,  find : 
(a)  The  circumference  and  the  area. 
(6)   The  side  and  area  of  the  inscribed  square, 
(c)  The  side  and  area  of  the  inscribed  regular  hexagon. 


234 


PLANE   GEOMETRY 


459.  PROBLEM.   To  inscribe  a  regular  decagon  in  a  given  circle. 

Given:  (?).     Required:  (?). 

Construction  :  Draw  any  radius 
AO.  Divide  it  into  mean  and  ex- 
treme ratio  (by  363),  having  the 
larger  segment  next  the  center. 
Take  A  as  a  center  and  OB  as  a 
radius,  draw  an  arc  cutting  O  at 
C.  Draw  AC,  BC,  OC. 

Statement  :  AC  is  a  side  of  the 
inscribed  regular  decagon.  Q.E.F. 

Proof:   AO  :  BO  =  BO  :  AB  (Const.). 

That  is,  AO  :  AC  =  AC  i  AB  (Ax.   6).     Hence,  A  ABC  and 
AOC  have  Z  A  common  and  are  similar  (?)  (317). 

.-.  1st,  Z  ACB  =  ZO  (?); 
and  2d,  A  ABC  is  isosceles  (being  similar  to  A  AOC). 

Hence,  AC=  BC(?),  but  AC  =  BO  (?).    .  '.BC=  BO  (Ax.  1). 

Therefore,  Z  BCO  =  Z  O  (?)  (55). 

.-.  Z  AGO  =  2  Z  O  (Ax.  2).     Also,  Z  A  =  Z  ACO  (?)  (55). 

/.Z  A      =  2ZO  (Ax.  1). 
Z  O      =  1  Z  Q.    Adding, 
A  of  A^CO  =  5Zo(Ax.  2). 

Hence,  5  y  O  =  180°  (?)  (110). 

.'.  Z  o  =  36°  (Ax.  3)  ;  that  is,  arc 
ference  Co  of  360°). 

Hence,  polygon  AE,  having  each  side 
regular  decagon  (?)  (418). 


C 


of  the  circum- 
is  an  inscribed 

Q.E.D. 


Ex.   If  the  radius  of  a  circle  is  20  in.,  find: 
(a)  The  circumference  and  area. 
(6)   The  side  and  area  of  the  inscribed  square, 
(c)   The  side  and  area  of  the  inscribed  regular  hexagon. 
(W)  The  side  of  the  inscribed  regular  decagon.     (See  365.) 
(e)  The  area  o.f  sector  AOC  (fig.  459). 

(/)  The  radius  of  a  circle  containing  twice  the  area  of  this 
circle. 


BOOK  V  235 

460.  PROBLEM.   To  inscribe  a  regular  is-gon  (pentedecagon)  in  a 
given  circle. 

Given:   (?).     Required:  (?). 

Construction :  Draw  AB,  the  side 
of  an  inscribed  hexagon,  and  AC, 
the  side  of  an  inscribed  decagon. 
Draw  BC. 

Statement :  BC  is  the  side  of  an 
inscribed  regular  15-gon.  Q.E.F. 

Proof:  Arc  BC=arc  AB  —  arc 
AC  =  i  —  J^  =  ^  of  the  circum- 
ference. (Const.) 

Hence,  the  polygon  having  each  side,  a  chord,  =  BC,  is  an 
inscribed  regular  15-gon  (?)  (418).  Q.E.D. 

461.  PROBLEM.    To  inscribe  in  a  given  circle: 

I.  A  regular  8-gon,  a  regular  i6-gon,  a  regular  32-gon,  etc. 
II.    A  regular  i2-gon,  24-gon,  etc. 

III.    A  regular  so-gon,  6o-gon,  etc. 

Construction  :  I.   Inscribe  a  square ;  bisect  the  arcs  ;  draw 
chords.     Statement:   (?).     Proof:   (?).     (See  422.)    Etc. 

II.  Inscribe  a  regular  hexagon ;  bisect  the  arcs.     Etc. 
III.  Inscribe  a  regular  15-gon,  etc. 

462.  PROBLEM.   To  inscribe  an  equilateral  triangle  in  a  circle. 

Construction:  Join  the  alternate  vertices  of  an   inscribed 
regular  hexagon.     Proof:  (?)     (See  421.) 

463.  PROBLEM.    To  inscribe  a  regular  pentagon  in  a  given  circle. 

464.  PROBLEM.   To  circumscribe  a  regular  polygon  about  a  circle. 

Construction :  Inscribe  a  polygon  having  the  same  num- 
ber of  sides.     At  the  several  vertices  draw  tangents. 

Statement:   (?).     Proof:  (?).     (See  419.) 


236 


PLANE   (GEOMETRY 


FORMULAS 
Sides  of  inscribed  polygons. 

1.  Side  of  inscribed  equilat- 
eral triangle  =  R  VS. 

Proof :  Z  ACS  is  a  rt.  Z  (?). 
AB  =  2  R  and  CB  =  R  (?)• 


2.  Side   of  inscribed  square 


Proof:  Use  fig.  of  457. 

3.  Side  of  inscribed  regular  hexagon  =  R  (?). 

4.  Side  of  inscribed  regular  decagon  =  %  R  (  V5  —  1).  (365.) 

466.  Sides  of  circumscribed  polygons. 

1.  Side   of   circumscribed 
equilateral  A  =  2  R  VS. 

Proof  :   Z  DAB  =  Z  DBA 
=  ZD  =  60°(?). 
.'.A  ABD  is  equilateral. 
AD  =  AB  =  R  _V3  (?). 
.'.DF=2R  V3. 

2.  Side   of    circumscribed 
square-  2  K  (?). 

3.  Side  of    circumscribed 
regular  hexagon  =  §  R  VS. 

(Explain.) 

467.  In  equilateral  triangle,  apothem  =  \  R. 
Proof:   Bisect  arc  AC  at  H.     Draw  OA,  OC,  ^!JJ,  CH. 
Figure  AOCH  is  a  rhombus.     (Explain.) 

ON  =  ±OH  =  ±R  (?)  (141). 


BOOK    V 


237 


468.  PROBLEM.   In  a  circle  whose  radius  is  /?  is  inscribed  a  regu- 
lar polygon  whose  side  is  s ;  to  find  the  formula  for  the  side  of  an 
inscribed  regular  polygon  having  double  the  number  of  sides. 

Given :  AB  =  s,  a  side  of  an  in- 
scribed regular  polygon  in  O 
whose  radius  is  It;  C,  the  mid- 
point of  arc  AB;  chord  AC. 

Required :  To  find  the  value  of 
AC,  the  side  of  a  regular  polygon 
having  double  the  number  of 
sides  and  inscribed  in  the  same 
circle. 

Construction:  Draw  radii  OA 
and  OC. 

Computation :   OC  bisects  AB  at  right  A  (?)  (70). 

In  rt.  A  AON,  O  is  an  acute  Z.     Hence  in  A  AOC, 
AC2  =  5I2  +  OC2  -  2  -  OC  -  ON  (?)  (346). 

But  AO  =  R,  OC  =  R,   ON  = 

or  AC 

469.  FORMULA.    If  R  =  1,  and  given  side  =  s,   the  side   of   a 
regular  polygon  having  twice  as  many  sides  =  \2-V4  —  s2. 


Ex.  1.   If  the  radius  of  a  circle  is  4,  find : 

(a)  The  side  of  the  inscribed  equilateral  triangle. 
(6)  The  side  of  the  circumscribed  equilateral  triangle, 
(c)  The  side  of  the  inscribed  square. 
(</)  The  side  of  the  circumscribed  square. 
(e)  The  side  of  the  inscribed  regular  hexagon. 
(/)  The  side  of  the  circumscribed  regular  hexagon. 
(</)  The  apothem  of  the  inscribed  equilateral  triangle. 
(A)  The  apothem  of  the  inscribed  regular  hexagon, 
(i)  The  side  of  the  inscribed  regular  dodecagon  (468). 
(j)  The  side  of  the  inscribed  regular  octagon. 


238  PLANE   GEOMETRY 

470.   PROBLEM.   To  find  the  approximate  numerical  value  of  TT. 
Given  :   A  circle  whose  diameter  =  D  and  circumference  =  C. 
Required  :  The  value  of  TT,  that  is,  the  value  of  C  -+•  I). 
Method:  1.  We  may  select  a  O  of  any  diameter   (442). 
Hence,  for  simplicity,  we  take  the  O  in  which  D=  2  ;  /.  R  =  1. 

2.  We  compute  the  perimeter  of  some  inscribed  regular 
polygon  (by  465). 

3.  We  compute  the  length   of   a  side   of   the   inscribed 
regular  polygon  having  double  the  number  of  sides,  by  the 

formula  s  =  \2—  A/4—  s2  (469).      From  this  we   can   find 
the  perimeter  of  this  polygon. 

4.  Using  the  side  of  this  polygon  as  known,  we  compute, 
by  the  same  formula,  a  side  of  the  inscribed  regular  polygon 
having  still  double  the  number  of  sides.      Hence  its  perim- 
eter can  be  found. 

5.  By  continuing  this  process  we  may  approximate  the 
value  of  the  circumference  (440,  I). 

6.  Thus  we  can  find  the  value  of  C  -f-  D,  or  TT. 
Computation  :   1.  Assume  E  =  1. 

2.  Consider  the  regular  hexagon  and  let  s6  represent  its 
side  and  PQ  its  perimeter.     Then  se=  1  and  P6=  6  (?). 

3.  Then,    a    side    of    the    inscribed   regular    12-gon     is 
*u  =  V2  ~  V^l  =  0.5176381,  and  pla  =  6.2116572. 

4.  Thus  we  may  find  *24  =  0.2610524  and  P24=  6.2652576. 

5.  By  continuing,  *8072  =  0.002045,  and  P3072  =  6.283184. 

6.  .-.  it  is  evident  that  C==  6.283184,  approximately. 

But,  TT=-  (?).    .-.  TT  =  6-288184  =  3.141592 +.    Q.E.F. 

D 


This  calculation  is 

tabulated  for  reference. 

s6  =  1,                .-.  P6 

=  6.                       s,92  =  0.032723, 

.-.  P192  =  6.282904. 

*12  =  0.517638,  .-.  P12 

=  6.211657.          Sgu  =  0.016362, 

.'•.  PM  =  6.283115. 

*24  =  0.261052,  .-.  P,< 

=  6.265257.          sm  =  0.008181, 

.-.  P768  =  6.283169. 

,4  =  0.130806,  .-.  P48 

=  6.278700.          *1686=  0.004091, 

...  p(rm  =  6.283180. 

s96  =  0.065438,  .-.  P,M 

=  6.282063.          53072=  0.002045, 

...  P^n  =  6.283184. 

BOOK   V  239 

ORIGINAL   EXERCISES   (NUMERICAL) 

MENSURATION  OF  REGULAR  POLYGONS  AND  THE  CIRCLE 

1.  Find  the  angle  and  the  central  angle  of: 

(i)  a  regular  pentagon ;  («)  a  regular  octagon ;    (m)    a  regular   do- 
decagon;  (i*y)  a  regular  20-gon. 

2.  Find  the  area  of  a  regular  hexagon  whose  side  is  8. 

3.  Find  the  area  of  a  regular  hexagon  whose  apothem  is  4. 

4.  In  a  circle  whose  radius  is  10  are  inscribed  an  equilateral  tri- 
angle, a  square,  and  a  regular  hexagon.     Find  the  perimeter,  apothem, 
and  area  of  each. 

5.  About  a  circle  whose  radius  is  10  are  circumscribed  an  equilat- 
eral triangle,  a  square,  and  a  regular  hexagon.     Find  the  perimeter  and 
area  of  each. 

6.  Find  the  circumference  and  area  of  a  circle  whose  radius  is  5 
inches.     [Use  TT  =  3^.] 

7.  Find  the  circumference  and  area  of  a  circle  whose  diameter  is 
42  centimeters. 

8.  The  radius  of  a  certain  circle  is  9  meters.    What  is  the  radius  of  a 
second  circle  whose  circumference  is  twice   as  long  as  the  first?   Of  a 
third  circle  whose  area  is  twice  as  great  as  the  first  ? 

9.  If  the  circumference  of  a  circle  is  55  yards,  what  is  its  diameter  ? 

10.  If  the  area  of  a  circle  is  113f  square  meters,  what  is  its  radius? 

11.  In  a  circle  whose  radius  is  35  there  is  a  sector  whose  angle  is  40°. 
Find  the  length  of  the  arc  and  the  area  of  the  sector. 

12.  The  area  of  a  circle   is  6^  times  the  area  of  another.      If  the 
radius  of  the  smaller  circle  is  12,  what  is  the  radius  of  the  larger  circle? 

13.  If  the  angle  of  a  sector  is  72°  and  its  arc  is  44  inches,  what  is 
the  radius  of  the  circle  ?  What  is  the  area  of  the  sector  ? 

14.  In  a  circle  whose  radius  is  7  find  the  area  of  the  segment  whose 
central  angle  is  120°.     [See  451.] 

15.  If  the  radius  of  a  circle  is  4  feet,  what  is  the  area  of  a  segment 
whose  arc  is  60°  ?  of  a  segment  whose  arc  is  a  quadrant? 

16.  Find  the  area  of  the  circle  inscribed  in  a  square  whose  area  is  75. 


240  PLANK    GEOMETRY 

17.  Find  the  area  of  an  equilateral  triangle  inscribed   in    a  circle 
whose  area  is  441  ir  square  meters. 

18.  If  the  length  of  a  quadrant  is  8  inches,  what  is  the  radius  ? 

19.  Find  the  length  of  an  arc  subtended  by  the  side  of  an  inscribed 
regular  15-gon  if  the  radius  is  4|  inches. 

20.  The  side  of  an  equilateral  triangle  is  10.    Find  the  areas  of  its  in- 
scribed and  circumscribed  circles. 

21.  Find  the  perimeter  and  area  of  a  segment  whose  chord  is  the 
side  of  an  inscribed  regular  hexagon,  if  the  radius  of  a  circle  is  5£. 

22.  A  circular  lake  9  rods  in  diameter  is  surrounded  by  a  walk  £  rod 
wide.     What  is  the  area  of  the  walk  ? 

23.  A  locomotive  driving  wheel   is  7  feet  in  diameter.     How  many 
revolutions  will  it  make  in  running  a  mile? 

24.  What  is  the  number  of  degrees  in  the  central  angle  whose  arc  is 
as  long  as  the  radius  ? 

25.  Find  the  side  of  the  square  equivalent  to  a  circle  whose  diameter 
is  4.2  meters. 

26.  Find  the  radius  of  that  circle  equivalent  to  a  square  whose  side 
is  5.5  inches. 

27.  Find  the  radius  of  the  circumference  which  divides  a  given  circle 
whose  radius  is  10£  into  two  equal  parts. 

28.  Three  equal  circles  are  each  tangent  to  the  other  two  and  the 
diameter  of  each  is  40  feet.     Find  the  area  between  these  circles. 

[Required  area  =  area  of  an  eq.  A  minus  area  of  three  sectors.] 

29.  Find  the  area  of  the  three  segments  of  a  circle  whose  radius  is 
5\/3,  formed  by  the  sides  of  the  inscribed  equilateral  triangle. 

30.  If  a  cistern  can  be  emptied  in  5  hours  by  a  2-inch  pipe,  how  long 
will  be  required  to  empty  it  by  a  1-inch  pipe  ? 

31.  Find  the  side,  apothem,  and  area  of  a  regular  decagon  inscribed 
in  a  circle  whose  radius  is  6  feet. 

32.  What  is  the  area  of  the  circle  circumscribed  about  an  equilateral 
triangle  whose  area  is  48  V3? 

33.  The  circumferences  of  two  concentric  circles  are  40  inches  and  50 
inches.     Find  the  area  of  the  circular  ring  between  them. 

34.  A  circle  has  an  area  of  80  square  feet.     Find  the  length  of  an 
arc  of  80°. 


HOOK    V 


35.  Find  the  angle  of  a  sector  whose  perimeter  equals  the  circum- 
ference. 

36.  Find  the  angle  of  a  sector  whose   area  is  equal  to  the  square 
of  the  radius. 

37.  Find  the  area  of  a  regular  octagon  inscribed  in  a  circle  whose 
radius  is  20. 

[Inscribe  square,  then  octagon.     Draw  radii  of  octagon.     Find  area  of 
one  isosceles  A  formed,  whose  altitude  is  half  the  side  of  the  square.] 

38.  A  rectangle  whose  length  is  double  its  width,  a  square,  an  equi- 
lateral triangle,  and  a  circle  all  have  the  same  perimeter,  namely  132 
meters.     Which  has  the  greatest  area?  the  least? 

39.  Through  a  point  without  a  circle  whose  radius  is  35  inches  two 
tangents  are  drawn,  forming  an  angle  of  60°.     Find  the  perimeter  and 
area  of  the  figure  bounded  by  the  tangents  and  their  smaller  intercepted  arc. 

40.  In  a  circle  whose  radius  is  12  are  two  parallel  chords  which  sub- 
tend arcs  of  60°  and  90°  respectively.     Find  the  perimeter  and  area  of 
the  figure  bounded  by  these  chords  and  their  intercepted  arcs. 

41.  A  quarter  mile  race  track  is  to  be  laid  out,  having  parallel  sides 
but  semicircular  ends  whose  radius  is  105  feet.     Find  the  length  of  the 
parallel  sides. 

42.  If  the  diameter  of  the  earth  is  3960  miles,  how  far  at  sea  can  the 
light  from  a  lighthouse  150  feet  high  be  seen  ? 

43.  The  diameter  of  a  circle  is  18  inches.     Find  the  area  of  the  figure 
between  this  circle  and  the  circumscribed  equilateral  triangle. 

44.  How  far  does  the  end  of    the   minute 
hand  of  a  clock  move  in  20  minutes,  if  the  hand 
is3|  inches  long? 

45.  The  diameter  of  a  circle   is    16   inches. 
What  is  the  area  of  that  portion  of  the  circle 
outside  the  inscribed  regular  hexagon  ? 

46.  Using  the  vertices  of  a  square  whose  side 
is  12,  as  centers,  and  radii  equal  to  4,  four  quad- 
rants are  described  within  the  square.     Find  the 
perimeter  and  area  of  the  figure  thus  formed. 

47.  Using  the  four  vertices  of  a  square,  whose 
side  is  12,  as   centers  and  radii  equal  to  6,  four 
arcs  are  described  without  the  square  (see  figure). 
Find  the  perimeter  and  area  of  the  figure  bounded 
by  these  four  arcs. 

BOBBINS'  PLANE  GEOM.  — 16 


242 


PLANE   GEOMETRY 


48.  Using  the  vertices  of  an  equilateral  triangle,  whose   side  is  16, 
as  centers  and  radii  equal  to  8,  three  arcs  are  described  within  the  tri- 
angle.    Find  the  perimeter  and  area  of  the  figure  bounded  by  these  arcs. 
Do  the  same  if  the  three  arcs  are  described  without  the  triangle  (terminat- 
ing in  the  sides,  in  each  case). 

49.  Using  the  vertices  of   a  regular  hexagon,  whose  side  is  20,  as 
centers  and  radii  equal  to  10,  six  arcs  are  described  within  the  hexagon. 
Find  the  perimeter  and  area  of  the  figure  bounded  by  these  arcs.     Do 
the  same  if  the  six  arcs  are  described  without  the  hexagon  (terminating 
in  the  sides,  in  each  case). 

50.  If  semicircumf  erences  be  described  within 
a  square,  whose  side  is  8  inches,  upon  the  four 
sides  as  diameters,  find  the  areas  of  the  four 
lobes  bounded  by  the  eight  quadrants.      Find 
the  area  of  any  one. 

In  the  following  exercises  let  n  =  the  number 
of  sides  of  the  regular  polygon;  .<?  =  the  length 
of  its  side  ;  r  =  its  apothein  ;  R  =  its  radius  ; 
K  =  its  area. 


51.    If  n  =  3,  show  that  s  =  R  V3;  r  =  |  R;  K  = 


52.  If  »  =  4,  show  that  s  =  R  V2  =  2r;  K  =  2R*  = 

53.  If  n  =  6,  show  that  s  =  R  =  2r^. 


54.    If  n  =  8,  show  that  s  =  R-^2  -V2=  2r(\/2-l);  r=—  JiTf  V2; 


-2  V2;  K=  2  R2  V2  =  8r2  (  \/2-  1). 


55.    If  n  =  10,  showthats  =  — 


56.  Ifw  =    5,  showthats  =  —  "^/lO  -  2  Vo;   r  =  —  (\/5+ 1). 

^  i 

57.  If  n  =  12  show  that  s  = 


BOOK  V  243 

58.  The  apothem  of  a  regular  hexagon  is  18  V3  inches.     Find  its 
side  and  area.     Find  the  area  of  the  circle  circumscribed  about  it. 

59.  What  is  the  radius  of  a  circle  whose  area  is  doubled  by  increas- 
ing the  radius  10  feet? 

60.  If  an  8-inch  pipe  will  fill  a  cistern  in  3  hours  20  minutes,  how 
long  will  it  require  a  2-inch  pipe  to  fill  it  ? 

61.  The  radius  of  a  circle  is  12  meters.     Find  : 
(a)    The  area  of  the  inscribed  square. 

(6)   The  area  of  the  inscribed  equilateral  triangle. 

(c)  The  area  of  the  inscribed  regular  hexagon. 

(d)  The  area  of  the  inscribed  regular  dodecagon. 

(e)  The  area  of  the  circumscribed  square. 

(/)  The  area  of  the  circumscribed  equilateral  triangle. 
(</)    The  area  of  the  circumscribed  regular  hexagon. 
Qi)    The  area  of  the  circumscribed  regular  dodecagon. 

62.  The  radius  of  a  circle  is  18.     Find  : 

(a)  The  side  and  apothem  of  the  inscribed  square. 

(6)  The  side  and  apothem  of  the  inscribed  equilateral  triangle. 

(c)  The  side  and  apothem  of  the  inscribed  regular  hexagon. 

(e?)  The  area  of  the  inscribed  square. 

(e)  The  area  of  the  inscribed  equilateral  triangle. 

(y)  The  area  of  the  inscribed  regular  hexagon. 

(g)  The  area  of  the  inscribed  regular  octagon. 

(h)  The  area  of  the  circumscribed  regular  hexagon. 

63.  Prove  that  the  area  of  an  inscribed  regular  hexagon  is  a  mean  pro- 
portional between   the   areas   of  the  inscribed  and  the   circumscribed 
equilateral  triangles.     [Find  the  three  areas  in  terms  of  -R.] 

64.  AB  is  one  side  of  an  inscribed  equilateral  triangle,  and  C  is  the 
midpoint  of  AB.     If  AB  be  prolonged  to  0  making  BO  equal  to  BC, 
and  OT  be  drawn  tangent  to  the  circle  at  T,  OT  will  be  f  the  radius. 

65.  A  square,  an  equilateral  triangle,  a  regular  hexagon,  and  a  circle 
all  have  the  same  area,  namely  5544  sq.  ft.     Which  figure  has  the  least 
perimeter  ?  the  greatest  ? 

66.  A  square,  an  equilateral  triangle,  a  regular  hexagon,  and  a  circle  all 
have  the  same  perimeter,  namely  396  in.     Find  their  areas  and  compare. 

67.  The  circumferences  of  two  concentric  circles  are  330  and  440  in. 
respectively.     Find  the  radius  of  another  circle  equivalent  to  the  ring 
between  these  two  circumferences. 


244  PLANE   GEOMETRY 

ORIGINAL   CONSTRUCTIONS 

1.  To  circumscribe  a  regular  hexagon  about  a  given  circle. 

2.  To  circumscribe  an  equilateral  triangle  about  a  given   circle. 

3.  To  circumscribe   a    regular  decagon    about   a    given    circle;    a 
regular  16-gon  ;  a  regular  24-gon  ;  a  square. 

4.  To   construct  an  angle  of  36°;  of  18°;  of  72°;  of  24°;  of  6°;  of 
48°;  of  96°. 

5.  To  construct  a  regular  hexagon  upon  a  given  line  as  a  side. 

6.  To  construct  a  regular  decagon  upon  a  given  line  as  a  side. 

7.  To  construct  a  regular  octagon  upon  a  given  line  as  a  side. 

8.  To  construct  a  regular  pentagon  upon  a  given  line  as  a  side. 

9.  To  construct  a  square  which  shall  have  double  the  area  of  a 
given  square. 

10.  To  inscribe  in  a  given 
circle  a  regular  polygon  simi- 
lar to  a  given  regular  polygon. 

Construction :  From  the 
center  of  the  polygon  draw 
radii.  At  the  center  of  the 
circle  construct  A  =  these  cen- 
tral  A  of  the  polygon.  Draw  chords.  Etc. 

11.  To  construct  a  regular  pentagon  which  shall  have  double  the  area 
of  a  given  regular  pentagon. 

12.  To  construct  a  circumference  equal  to  the  sum  of  two  given  cir- 
cumferences. 

13.  To  construct  a  circumference  which  shall  be  three  times  a  given 
circumference. 

14.  To  construct  a  circumference  equal  to  the  difference  of  two  given 
circumferences. 

15.  To  construct  a  circle  whose  area  shall  be  five  times  a  given  circle. 

16.  To  construct  a  circle  equivalent  to  the  sum  of  two  given  circles ; 
another,  equivalent  to  their  difference. 

17.  To  construct  a  circle  whose  area  shall  be  half  a  given  circle. 

18.  To  bisect  the  area  of  a  given  circle  by  a  concentric  circumference. 

19.  To  divide  a  given  circumference  into  two  parts  which  shall  be  in 
the  ratio  of  3:7;  into  two  other  parts  which  shall  be  in  the  ratio  of 
5:7;  into  still  two  other  parts,  in  the  ratio  of  8  : 7. 


HOOK    V 


245 


MAXIMA     VXD   MINIMA 

471.  Of   geometrical    magnitudes  which    satisfy   a   given 
condition   (or  given  conditions)    the  greatest  is  maximum, 
and  the  least  is  minimum. 

Thus,  of  all  chords  that  can  be  drawn  through  a  given  point  within 
a  circle,  the  diameter  is  the  maximum,  and  the  chord  perpendicular  to 
the  diameter  at  the  point  is  the  minimum. 

Isoperimetric  figures  are  figures  having  equal  perimeters. 

472.  THEOREM.    Of  all  triangles  having  two  given  sides,  that  in 
which  these  sides  form  a  right  angle  is  the  maximum. 

Given  :    A  ABC  and  A  ABD  c 

having     AB     common,     and 
AC  =  AD,  but  Z  CAB  a  rt.  Z 
and  /.DAB  not  a  right  Z. 
To  Prove:  A  ABC  >  A  ABD. 
Proof  :    Draw  altitude  DE. 
Now  AD  >  !)#(?).      /.  AC  >  DE  (Ax.    6). 
Multiply  each  member  by  ^  AB. 
Then  J  AB  -  AC  >  J  AB  •  DE  (?). 
Now  J  AB  -  AC  —  area  A  ABC  (?), 
and        i  AB  -  DE  =  area  A  ABD  (?). 

Therefore,  A  ABC  >  A  ABD  (Ax.  6).        Q.E.D. 

This  theorem  may  be  stated  thus  :  Of  all  triangles  having 
two  given  sides,  that  triangle  whose  third  side  is  the  diameter 
of  the  circle  which  circumscribes  it  is  the  maximum. 

Therefore,  Of  all  n-gons  having  n  —\  sides  given,  that  poly- 
gon whose  nth  side  is  the  diameter  of  a  circle  which  circum- 
scribes the  polygon  is  the  maximum. 

Ex.  1.  Of  all  parallelograms  having  two  adjacent  sides  given,  the 
rectangle  is  the  maximum. 

Ex.  2.  Of  all  lines  that  can  be  drawn  from  an  external  point  to  a  cir- 
cumference, which  is  the  maximum  ?  the  minimum  V 


246 


PLANE   GEOMETRY 


,,E 
: 


473.  THEOREM.  Of  all  isoperimetric  triangles  having  the  same 
base  the  isosceles  triangle  is  the  maximum. 

Given  :    A  ABC  and  ABD 

isoperimetric,  having  the 
same  base,  AB,  and  A  ABC 
isosceles. 

To  Prove  : 
A  ABC  >  A  ABD. 
Proof  :    Prolong  AC  to  E, 
making  CE  =  AC,  and  draw 
BE.      Using  D  as  a   center 
and  BD  as  a  radius,  describe 
an  arc  cutting  EB  prolonged, 
at  F.    Draw  CG  and  DH  \\  to 
AB,  meeting  EF  at  G  and  //  respectively.     Draw  AF. 

Now,  using  C  as  a  center  and  AC  or  BC  or  EC  as  a  radius, 
the  circle  described  will  pass  through  A,  B,  and  E  (Hyp. 
and  Const.).  .'.  /.  ABE  =  rt.  Z  (?). 

That  is,  AB  is  J_  to  EF.     Hence,  CG  and  DH  are  _L  to  EF  (?). 
AC  +  CE  =  AC  +  CB  =  AD  +  DB  =  AD  +  DF  (Hyp.   and 
Const,). 

That  is,  AE  =  AD  +  DF  (Ax.    1). 
But  AD  +  DF  >  AF  (?).      .'.  AE  >  ^LF  (Ax.  6). 
.-.  BE  >  BF  (?)   (90),  and  J  BE  >  £  BF  (?). 
Now,  BG  =  ^  BE  and  £//  =  J  #F  (?)  (73,  Cor.). 
.'.  BG  >  7?ff  (Ax.  6). 
Multiply  each  member  by  |-  AB. 
Then,  |  ^is  •  BG  >  J  ^J5  •  BII  (?). 
But,     |  AB  -  BG  =  area  A^BC  (?), 
and          I  AB  •  BH  =  area  A  ^BD  (?). 

.'.A  ABC  >  A  ABD   (Ax.  6).  Q.E.D. 

474.  THEOREM.  Of  isoperimetric  triangles  the  equilateral  tiiangle 
is  the  maximum. 

[Any  side  may  be  considered  the  base.] 


BOOK  V  247 

475.  THEOREM.    Of  isoperimetric  polygons  having  the  same  num- 
ber of  sides  the  maximum  is  equilateral. 

Given  :    Polygon  AD,  the  max- 
imum   of    all    polygons    having  /   *''\ \-M 

the  same  perimeter  and  the  same 
number  of  sides. 

To   Prove  :     AB  —  BC  =  CD  =  F< 
DE  =  etc. 

Proof  :    Draw  AC  and  suppose 
AB  not  =  BC. 

On    AC    as     base,     construct 
A  ACM  isoperimetric  with  A  ABC  and  isosceles  ;  that  is,  make 
AM=CM.     Then  A  ACM  >  A  ABC   (?)  (473). 

Add  to  each  member,  the  polygon  ACDEF. 

.*.  polygon  AMCDEF  >  polygon  AD  (?). 

But  the  polygon  AD  is  maximum  (Hyp.). 

.*.  AB  cannot  be  unequal  to  BC  as  we  supposed  (because 
that  results  in  an  impossible  conclusion). 

Hence,  AB  =  BC.    Likewise  it  is  proved  that  BC  =  CD  =  etc. 

Q.E.D. 

476.  THEOREM.    Of  isoperimetric  polygons  having  the  same  num- 
ber of  sides  the  equilateral  polygon  is  maximum. 

Proof  :  Only  one  such  polygon  is  maximum,  and  the  maxi- 
mum is  equilateral  (475). 

Only  one  such  polygon  is  equilateral,  hence  the  equilateral 
polygon  and  the  maximum  polygon  are  the  same.  Q.E.D. 


Ex.  1.   Of  isoperimetric  triangles,  the  maximum  is  equilateral. 

Ex.  2.  Of  all  right  triangles  that  can  be  constructed  upon  a  given 
hypotenuse,  which  is  maximum?  Why  ? 

Ex.  3.  Of  all  triangles  having  a  given  base  and  a  given  vertex-angle, 
the  isosceles  is  the  maximum. 

Ex.  4.  Of  all  mutually  equilateral  polygons,  that  which  can  be  in- 
scribed in  a  circle  is  the  maximum. 


248 


PLANE   GEOMETRY 


477.  THEOREM,  Of   isoperimetric  regular  polygons,   the   polygon 
having  the  greatest  number  of  sides  is  maximum. 

Given :  Equilateral  A  ABC 
and  square  /S,  having  the  same 
perimeter. 

To  Prove :  Square  S>AABC. 

Proof :  Take  D,  any  point  in 
J5C,  and  draw  AD.  On  AD  as 
base,  construct  isosceles  A  ADE,  isoperimetric  wh\iAABD. 

Now  A  AED  >  A  ABD  (?)  (473). 

Adding  A  ADC  to  each  member,  AEDC  >  A  ABC  (?). 

AEDC  is  isoperimetric  with  A  ABC  find  S  (Hyp.  and  Const.). 

Hence,  S  >  AEDC  (?)  (476). 

Therefore  8  >  A  ABC  (?)  (Ax.  11). 

Similarly  we  may  prove  that  an  isoperimetric  regular  pen- 
tagon is  greater  than  8 ;  and  an  isoperimetric  regular  hexa- 
gon is  greater  than  this  pentagon,  etc. 

Therefore,  the  regular  polygon  having  the  greatest  num- 
ber of  sides  is  maximum.  Q.E.D. 

478.  THEOREM.    Of  all  isoperimetric  plane  figures  the  circle  is  the 
maximum, 

479.  THEOREM.    Of  equivalent  regular  polygons  the  perimeter  of 
the  polygon  having  the  greatest  number  of  sides  is  the  minimum. 


Given :   Any  two  equivalent  regular  polygons,  A  and  -B,  A 
having  the  greater  number  of  sides. 


BOOK  V  249 

To  Prove :  The  perimeter  of  A  <  the  perimeter  of  B. 
Proof :  Construct    regular    polygon    s,   similar  to  B  and 
isoperimetric  with  A. 

Then  A  >  S  (477),  but  A  =0=  B  (?).    .-.  B  >  s  (?)  (Ax.  6). 

Hence,  the  perimeter  of  B  >  perimeter  of  s  (390). 

But,  the  perimeter  of  8  =  perimeter  of  A  (?). 

.'.  perimeter  of  B  >  perimeter  of  A  (Ax.  6). 

That  is,  the  perimeter  of  A<  the  perimeter  of  B.       Q.E.D. 

480.  THEOREM.   Of  all  equivalent  plane  figures  the  circle  has  the 
minimum  perimeter. 

ORIGINAL   EXERCISES 

1.  Of  all  equivalent  parallelograms  having  equal  bases  the  rectangle 
has  the  minimum  perimeter. 

2.  Of  all  lines  drawn  between  two  given  parallels  (terminating  both 
ways  in  the  parallels),  which  is  the  minimum  ?    Prove. 

3.  Of  all  straight  lines  that  can  be  drawn  on  the  ceiling  of  a  room 
12  feet  long  and  9  feet  wide,  what  is  the  length  of  the  maximum? 

4.  Find  the  areas  of  an  equilateral  triangle,  a  square,  a  regular  hex- 
agon, and  a  circle,  the  perimeter  of  each  being  264  inches.     Which  is 
maximum?     What  theorem  does  this  exercise  illustrate  ? 

5.  Find  the  perimeters  of  an  equilateral  triangle,  a  square,  a  regular 
hexagon,  and  a  circle,  if  the  area  of  each  is  1386  square  feet.     Which 
perimeter  is  the  minimum?     What  theorem  does  this  exercise  illustrate? 

6.  Of  isoperimetric  rectangles  which  is  maximum? 

7.  To  divide  a  given  line  into  two  parts  such 
that  their  product  (rectangle)  is  maximum. 

8.  Of  all    equivalent  triangles   having   the 
same  base  the  isosceles  triangle  has  the  minimum 
perimeter. 

To  Prove:    The   perimeter   of   &ABC  >  the 
perimeter  of  A  AB'C. 

Proof:  AD<AB'  +  B'C;  etc.  A 

9.  Of  all  rectangles  inscribed  in  a  circle  which  is  maximum  ?  Prove. 

10.  Of  all  rectangles  inscribed  in  a  semicircle  which  is  maximum? 
Prove. 

11.  Of  all  equivalent  rectangles,  the  square  has  the  minimum  per- 
imeter. 


250  PLANE    GEOMETRY 

12.  Of  all  triangles  having  a  given  base  and  a  given  vertex-angle,  the 
isosceles  triangle  has  the  maximum  perimeter. 

13.  Of  all  triangles  having  a  given  altitude  and  a  given  vertex-angle, 
the  isosceles  triangle  is  the  minimum. 

14.  Of  all  triangles  that  can  be  inscribed  in  a  given  circle  the  equi- 
lateral triangle  has  the  maximum  area  and  the  maximum  perimeter. 

15.  The  cross  section  of  a  bee's  cell  is  a  regular  hexagon.    Would  this 
be  the  most  economical  for  the  bee,  if  one  cell  in  a  hive  were  all  he  were 
to  fill  (that  is,  would  he  use  the  least  wax)  ?    Considering  also  the  adjoin- 
ing cells,  does  the  form  of  the  regular  hexagon  require  the  least  wax? 
Explain.     Does  it  also  permit  the  storing  of  the  most  honey  ?    Why? 

16.  Prove  that  a  regular  hexagon  is  greater  than  an  isoperimetric 
square,  by  the  method  employed  in  477. 

17.  Answer  the  questions  of  exercise  65  on  page  243,  without  any 
computation.     Give  reasons. 

18.  Compare  the  areas  of  the  figures  mentioned  in  exercise  66,  page 
243,  without  performing  any  computation. 


INDEX   OF   DEFINITIONS 


(The  numbers  refer  to  pages.) 


\bbreviations,  16. 
Acute  angle,  13. 
Acute  triangle,  15. 
Adjacent  angles,  12. 
Adjacent-interior  angles,  38. 
Adjoining-interior  angles,  38. 
Alternate  angles,  38. 
Alternate-exterior  angles,  38. 
Alternate-interior  angles,  38. 
Alternation,  141. 
Altitude,  of  parallelogram,  46. 

of  trapezoid,  46. 

of  triangle,  33. 
Analysis,  127. 
Angle,  12. 

acute,  13. 

bisector  of,  14,  33. 

central,  80. 

central,  of  regular  polygon,  220. 

complement  of,  13. 

degree  of,  13. 

exterior,  of  polygon,  54. 

exterior,  of  triangle,  42. 

inscribed,  in  circle,  80. 

inscribed,  in  segment,  102. 

measure  of,  102. 

oblique,  13. 

obtuse,  13. 

plane,  12. 

right,  12. 

sides  of,  12. 

straight,  13. 

supplement  of,  13. 

vertex  of,  12. 
Angles,  adjacent,  12. 

adjacent-interior,  38. 

adjoining-interior,  38. 

alternate,  38. 

alternate-exterior,  38. 

alternate-interior,  38. 

complementary,  13. 

corresponding,  38. 

equal,  15. 

homologous,  of  polygons,  55. 


Angles  —  Continued 

included,  14. 

interior,  38. 

of  polygons,  54. 

of  quadrilateral,  45. 

of  triangle,  14. 

opposite-interior,  42. 

supplementary,  13. 

vertical,  12. 
Antecedents,  140. 
Apothem,  220. 
Arc,  80. 

degree  of,  102. 

intercepted,  81. 

subtended,  81. 
Arcs,  similar,  228. 
Area,  187. 
Auxiliary  lines,  25. 
Axiom,  16. 

parallel,  36. 
Axioms,  16. 
Axis  of  symmetry,  58. 

Base,  of  figure,  45. 

of  isosceles  triangle,  15. 

of  triangle,  14. 
Bases,  of  parallelogram,  45. 

of  trapezoid,  45. 
Bisector  of  angle,  14,  33. 
Boundary,  12. 

Center,    figure    symmetrical    with    re- 
spect to,  58. 

of  circle,  79. 

of  circumference,  79. 

of  regular  polygon,  220. 

of  symmetry,  58. 
Central  angle,  80. 

of  regular  polygon,  220. 
Chord,  79. 
Circle,  79. 

angle  inscribed  in,  80. 

center  of,  79. 

circumscribed  about  polygon,  91. 

251 


INDEX   OF   DEFINITIONS 


Circle  —  Continued 

diameter  of,  79. 

inscribed  in  polygon,  92. 

radius  of,  79. 

sector  of,  80. 

segment  of,  80. 

tangent  to,  79. 
Circles,  concentric,  80. 

equal,  80. 

escribed,  124. 

tangent  externally,  80. 

tangent  internally,  80. 
Circumference,  79. 

center  of,  79. 
Circumscribed  circle,  91. 
Circumscribed  polygon,  92. 
Commensurable  quantities,  97 
Common  tangent,  92. 
Complement  of  angle,  13. 
Complementary  angles,  13. 
Composition,  142. 
Composition  and  division,  143. 
Concave  polygon,  54. 
Concentric  circles,  80. 
Conclusion,  23. 
Concurrent  lines,  57. 
Consequents,  140. 
Constant,  97. 
Construction,  115. 
Continued  proportion,  140, 
Converse  of  a  theorem,  25. 
Convex  polygon,  54. 
Corollary,  17. 
Corresponding  angles,  38. 
Curved  line,  79. 

Decagon,  55. 
Degree,  of  angle,  13. 

of  arc,  102. 
Demonstration,  17. 

elements  of,  23. 
Determination,  of  a  straight  line,  11. 

of  a  circle,  116. 
Diagonal,  45. 
Diameter,  79. 
Direct  proportion,  160. 
Discussion,  115. 
Distance,  between  two  points,  25. 

from  a  point  to  a  line,  32. 
Division,  142. 
Dodecagon,  55. 

Equal  angles,  15. 
Equal  circles,  80 


Equal  figures,  15. 
Equal  polygons,  55. 
Equiangular  polygon,  54. 
Equiangular  triangle,  15. 
Equilateral  polygon,  54. 
Equilateral  triangle,  15. 
Equivalent  figures,  187. 
Escribed  circles,  124. 
Exterior  angle,  of  a  polygon,  54. 

of  a  triangle,  42. 
Extreme  and  mean  ratio,  182. 
Extremes,  140. 

Figure,  base  of,  45. 

rectilinear,  11. 

symmetrical  with  respect  to  center, 
58. 

symmetrical    with  respect    to    line, 

58. 
Figures,  equal,  15. 

equivalent,  187. 

isoperimetric,  245. 
Foot  of  perpendicular,  13. 
Fourth  proportional,  140. 

Geometry,  11. 
Plane,  11. 

Harmonic  division,  149. 

Heptagon,  55. 

Hexagon,  55. 

Homologous  angles  in  polygons,  55. 

Homologous  parts,  15. 

Homologous  sides,  in  polygons,  55. 

in  triangles,  155. 
Hypotenuse,  15. 
Hypothesis,  23. 

Included  angles,  14. 
Incommensurable  quantities,  97. 
Indirect  method,  36. 
Inscribed  angle,  in  circle,  80. 

in  segment,  102. 
Inscribed  circle,  92. 
Inscribed  polygon,  91. 
Intercept,  to,  51,  81. 
Intercepted  arc,  81. 
Interior  angles,  38. 
Intersect,  to,  51. 
Inverse  proportion,  160. 
Inversion,  142. 
Isoperimetric  figures,  245. 
Isosceles  trapezoid,  45. 
Isosceles  triangle,  15. 


INDEX  OF   DEFINITIONS 


253 


Legs,  of  isosceles  trapezoid,  45. 

of  isosceles  triangle,  15. 
Legs  of  right  triangle,  15. 
Limit  of  variable,  97,  99. 
Limits,  theorem  of,  99. 
Line,  11. 

curved,  79. 

divided  harmonically,  149. 

divided    into    extreme    and    mean 
ratio,  182. 

tangent  to  circle,  79. 
Lines,  auxiliary,  25. 

concurrent,  57. 

divided  proportionally,  147. 

oblique,  13. 

parallel,  36. 

perpendicular,  13. 
Locus,  60. 

Maximum,  245. 

Mean  proportional,  140. 

Means,  140. 

Measure,  of  angle,  102. 

of  quantity,  97. 

unit  of,  97. 
Median,  of  trapezoid,  46. 

of  triangle,  33. 
Method,  indirect,  36. 

of  exclusion,  36. 
Minimum,  245. 
Motion,  12. 

Mutually  equiangular  polygons,  54. 
Mutually  equilateral  polygons,  54. 

N-gon,  55. 

Oblique  angle,  13. 
Oblique  lines,  13. 
Obtuse  angle,  13. 
Obtuse  triangle,  15. 
Octagon,  55. 
Opposite  interior  angles,  42. 

Parallel  axiom,  36. 
Parallel  lines,  36. 
Parallelogram,  45. 

altitude  of,  46. 

bases  of,  45. 
Pentagon,  55. 
Pentedecagon,  55. 
Perimeter,  92. 
Perpendicular,  13. 

foot  of,  13. 
Pi  (TT),  227. 


Plane,  11. 
Plane  angle,  13. 
Plane  Geometry,  11. 
Point,  11. 

equally  distant  from  two  lines,  32. 

of  contact,  79. 

Points,  symmetrical  with  respect  to  a 
line,  58. 

symmetrical  with  respect  to  a  point, 

58. 
Polygon,  54. 

angles  of,  54. 

center  of  regular,  220. 

central  angle  of  regular,  220. 

circumscribed  about  a  circle,  92. 

concave,  54. 

convex,  54. 

equiangular,  54. 

equilateral,  54. 

exterior  angle  of,  54. 

inscribed  in  a  circle,  91. 

re-entrant,  54. 

regular,  217. 

vertices  of,  54. 
Polygons,  equal,  55. 

mutually  equiangular,  54. 

mutually  equilateral,  54. 

similar,  150. 
Postulate,  17. 
Problem,  115. 
Projection,  of  a  line,  163. 

of  a  point,  163. 
Proof,  17. 
Proportion,  140. 

antecedents  of,  140. 

consequents  of,  140. 

continued,  140. 

direct,  160. 

extremes  of,  140. 

inverse,  160. 

means  of,  140. 

reciprocal,  160. 
Proportional,  fourth,  140. 

mean,  140. 

third,  140. 
Proposition,  115. 

Q.E.D.,  23. 
Q.E.F.,  116. 
Quadrant,  80. 
Quadrilateral,  45. 

angles  of,  45. 

sides  of,  45. 

vertices  of,  45. 


254 


INDEX   OF    DEFINITIONS 


Quantities,  commensurable,  97. 
incommensurable,  97. 

Radius,  of  circle,  79. 

of  regular  polygon,  220. 
Ratio,  97,  140. 

extreme  and  mean,  182. 

series  of  equal,  140. 
Reciprocal  proportion,  160. 
Rectangle,  45. 
Rectilinear  figure,  11. 
Reductio  ad  Absurdum,  36. 
Re-entrant  polygon,  54. 
Regular  polygon,  217. 
Rhomboid,  45. 
Rhombus,  45. 
Right  angle,  12. 
Right  triangle,  15. 

Scalene  triangle,  15. 
Secant,  79. 
Sector  of  circle,  80. 
Sectors,  similar,  228. 
Segment  of  circle,  80. 
Segments,  of  line,  148. 

similar,  228. 
Semicircle,  80. 
Semicircumference,  80. 
Series  of  equal  ratios,  140. 
Sides,  homologous,  in  polygons,  55. 

homologous,  in  triangles,  155. 

of  angle,  12. 

of  polygon,  54. 

of  quadrilateral,  45. 

of  triangle,  14. 
Similar  arcs,  228. 
Similar  polygons,  150. 
Similar  sectors,  228. 
Similar  segments,  228. 
Similar  triangles,  150. 
Solid,  11. 
Square,  45. 
Statement,  115. 
Straight  angle,  13. 
Straight  line,  11. 

divided  harmonically,  149. 

divided    into    extreme    and    mean 
ratio,  182. 

tangent  to  a  circle,  79. 
Straight  lines  divided  proportionally. 

147. 

Subtend,  to,  80. 
Subtended  arc,  81. 
Superposition,  15. 


Supplement  of  an  angle,  13. 
Supplementary  angles,  13. 
Surface,  11. 
Symbols,  16. 
Symmetry,  58. 

axis  of,  58. 

center  of,  58. 

Tangent,  79. 

circles,  80. 
Theorem,  17. 

converse  of,  25. 

elements  of,  23. 

of  limits,  99. 
Third  proportional,  140. 
Transversal,  38. 
Trapezium,  45. 
Trapezoid,  45. 

altitude  of,  46. 

bases  of,  45. 

isosceles,  45. 

legs  of,  45. 

median  of,  46. 
Triangle,  14. 

acute,  15. 

altitude  of,  33. 

angles  of,  14. 

base  of,  14. 

equiangular,  15. 

equilateral,  15. 

exterior  angle  of,  42. 

isosceles,  15. 

median  of,  33. 

obtuse,  15. 

right,  15. 

scalene,  15. 

sides  of,  14. 

vertex  of,  14. 

vertices  of,  14. 
Triangles,  similar,  150. 

Unit,  of  measure,  97. 
of  surface,  187. 

Variable,  97. 

limit  of,  97,  99. 
Vertex,  of  an  angle,  12. 

of  a  triangle,  14. 
Vertex-angle,  14. 
Vertical  angles,  12. 
Vertices,  of  polygon,  54. 

of  quadrilateral,  45. 

of  triangle,  14. 


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